blog-title#highlight" class="m-post-header__title"> Using The "Definite Assignment Assertion" To Define Required Input Bindings In Angular 7.1.1

In an Angular application, Directives represent a special type of Class. In a sense, they are a standard JavaScript Class with properties and methods. But, unlike a Class that gets created and consumed explicitly in your own logic, Directive classes are created, consumed, and managed implicitly by the Angular renderer. As such, properties that might otherwise be required as a constructor argument are, instead, provided through template input bindings. Which begs the question: how do you define a "required" property / input binding that is not actually available at the time of instantiation? As explained in the strictPropertyInitialization flag clean-up issue on GitHub , people on the Angular team have started to use TypeScript's "Definite Assignment Assertion" to denote required Class properties as "defined" even when they won't truly be defined until after the Directive's class constructor has been executed. I had never seen the "Definite Assignment Assertion" before, so I wanted to try it out for myself in a Component Directive in Angular 7.1.1.

Run this demo in my JavaScript Demos project on GitHub .

View this code in my JavaScript Demos project on GitHub .

To explore this concept, I created a "Badge" component that accepts a [user] input binding and renders the badge template with name, email, and avatar. The whole reason-to-be for this component is to render the passed-in User object. As such, this component makes no sense without the [user] input. This input binding is therefore required in order for this component to work. Without the user property, this Badge component would be in an "invalid state".

Before learning about the "Definite Assignment Assertion" notation in TypeScript, I might have tried to approach this Component definition by making the "user" property optional (meaning, for example, that it can also be set to null):

In this case, in the BadgeComponent constructor, I have to set the user property to null otherwise TypeScript complains that I didn't fully initialize the class property (thanks to the "strictPropertyInitialization" tsconfig setting). Of course, in the ngOnInit() and ngOnChanges() event-handlers, I am attempting to assert that the value must be defined (as a required input binding). But, when I go to compile this code, I get the following TypeScript error:

ERROR in app/badge.component.ts.BadgeComponent.html(4,10): : Object is possibly 'null'.

As you can see, TypeScript and Angular are complaining that I am attempting to use a potentially null value in my template expressions. Now, all of the logic for this component happens to be in the template; but, if my Class made additional references to the "user" property within its methods, similar errors would be thrown by the TypeScript compiler.

To fix this workflow, Angular internals have started to use the "Definite Assignment Assertion". This is a property name followed by an exclamation point. For example:

public user! : User;

This assertion tells TypeScript that the "user" property (aka, our required input binding) may look like it's not being fully initialized; but, that TypeScript should trust that the application is going to define this value before it is consumed by the given Class. In other words, we are telling TypeScript to treat this as a required property, but to not validate its initialization.

Of course, TypeScript can only make assertions about compile-time truths - it doesn't know anything about the runtime. As such, this won't prevent runtime errors if the "user" property is undefined; but, this assertion will better express the intended use of the Component Class.

Let's take a look at what the BadgeComponent looks like when we add the "Definite Assignment Assertion" notation:

As you can see, we are denoting the "user" property as "defined". Which allows the code to compile even though we've commented-out the constructor initialization. In this case, I'm still using the ngOnInit() and ngOnChanges() life-cycle methods to validate the input binding. But these are here to better report runtime errors and express the intent of the class. After all, if the input binding is missing, something is going to break.

To see this "Definite Assignment Assertion" in action, I've created an App component that attempts to consume the BadgeComponent both with and without a provided [user] input binding:

As you can see, the last of the "bn-badge" elements is omitting the [user] input binding. And, when we run this code in the browser, we get the following output:

The Definite Assignment Assertion in TypeScript makes it easy to denote Angular Directive input bindings as required without having to initialize them.

Clearly, if the [user] input binding is required but omitted from the calling code, the page will still break. After all, the BadgeComponent is in an invalid state. But, the BadgeComponent itself expresses clear intent as to how the input binging is supposed to be used.

When you author an Angular Component, some input bindings are optional and some are required. With optional input bindings, you can provide default or fallback values in order to make the class properties easier to work with. But, when it comes to required input bindings, there is no default or fallback. As such, TypeScript will complain that the required properties aren't properly initialized. Thankfully, the "Definite Assignment Assertion" allows us to tell TypeScript to relax the property validation and trust that the Angular application will define these values before they are used.

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Reader Comments

angular add definite assignment assertion to property

Hello, Nice Article. I am actually trying to utilize the definite assertion in my angular project. I have a model class of my object, which is your interface by the way. But Im trying to model the db. So i have a class Rating{rating:number;} And an export class Review which has rating!:Rating; In my ts class, i did review.rating.rating = 0, but it is seeing review.rating as undefined.

Please can you help me

angular add definite assignment assertion to property

So, the Definite Assignment Assertion won't actually influence the runtime value of the property - it will only prevent the TypeScript compiler from complaining that the property isn't being initialized. The point of the Definite Assignment Assertion is handle cases in which a property is being set outside of the current context (such as with an input-binding), and we are assuring the compiler that this value will be set by the time it is consumed.

Based on what you are saying, it sounds like you are attempting to reference reviewer.rating before you actually set it.

I believe in love . I believe in compassion . I believe in human rights . I believe that we can afford to give more of these gifts to the world around us because it costs us nothing to be decent and kind and understanding. And, I want you to know that when you land on this site, you are accepted for who you are , no matter how you identify, what truths you live, or whatever kind of goofy shit makes you feel alive! Rock on with your bad self!

TypeScript definite assignment assertions

TypeScript never stops improving, although most changes over the past year have been “non syntactical” – i.e. there have been a huge swathe of improvements to how types are handled, and a large slice of improvements to make the tooling even better. It has a been a while, though, since we got a new character to decorate our code. The wait is over, thanks to the TypeScript Definite Assignment Assertion. Let’s take a look at it with a short example.

Warning Triangle

No definite assignment

The new feature is related to the following improved compile-time check. In the example below, I forgot to assign a value to the wordsPerMinute property. This can happen when you forget to add a default value, or when you forget to initialize it in the constructor, or (as below) when you forget to map a parameter to the property (remember, you don’t need to manually map constructor parameters !).

Whatever the reason, if you compile using the strict flag (I keep telling you to use it), you’ll get the following error, known as a definite assignment error because there is no definite assignment:

app.ts(2,13): error TS2564: Property ‘wordsPerMinute’ has no initializer and is not definitely assigned in the constructor.

Fixing, and definite assignment assertions

The correct fix is probably to assign this.wordsPerMinute = wordsPerMinute in the constructor – but in some cases, you may be doing something funky where the dependency will be resolved in a way the compiler is unable to determine.

When you need to allow a property with no definite assignment, you can use the definite assignment assertion . This is a very grand name for adding a bang (!) to the property name.

This will only work in TypeScript 2.7 and newer.

On the whole, unless you have an iron-clad reason to use it – you’ll probably want to avoid the definite assignment assertion. In most cases, the real value of this feature lies in the part that detects unassigned properties.

Warning Triangle, Public Domain. Wikipedia .

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TypeScript 2.7

Constant-named properties.

TypeScript 2.7 adds support for declaring const-named properties on types including ECMAScript symbols.

This also applies to numeric and string literals.

unique symbol

To enable treating symbols as unique literals a new type unique symbol is available. unique symbol is a subtype of symbol , and are produced only from calling Symbol() or Symbol.for() , or from explicit type annotations. The new type is only allowed on const declarations and readonly static properties, and in order to reference a specific unique symbol, you’ll have to use the typeof operator. Each reference to a unique symbol implies a completely unique identity that’s tied to a given declaration.

Because each unique symbol has a completely separate identity, no two unique symbol types are assignable or comparable to each other.

Strict Class Initialization

TypeScript 2.7 introduces a new flag called strictPropertyInitialization . This flag performs checks to ensure that each instance property of a class gets initialized in the constructor body, or by a property initializer. For example

In the above, if we truly meant for baz to potentially be undefined , we should have declared it with the type boolean | undefined .

There are certain scenarios where properties can be initialized indirectly (perhaps by a helper method or dependency injection library), in which case you can use the new definite assignment assertion modifiers for your properties (discussed below).

Keep in mind that strictPropertyInitialization will be turned on along with other strict mode flags, which can impact your project. You can set the strictPropertyInitialization setting to false in your tsconfig.json ’s compilerOptions , or --strictPropertyInitialization false on the command line to turn off this checking.

Definite Assignment Assertions

The definite assignment assertion is a feature that allows a ! to be placed after instance property and variable declarations to relay to TypeScript that a variable is indeed assigned for all intents and purposes, even if TypeScript’s analyses cannot detect so.

With definite assignment assertions, we can assert that x is really assigned by appending an ! to its declaration:

In a sense, the definite assignment assertion operator is the dual of the non-null assertion operator (in which expressions are post-fixed with a ! ), which we could also have used in the example.

In our example, we knew that all uses of x would be initialized so it makes more sense to use definite assignment assertions than non-null assertions.

Fixed Length Tuples

In TypeScript 2.6 and earlier, [number, string, string] was considered a subtype of [number, string] . This was motivated by TypeScript’s structural nature; the first and second elements of a [number, string, string] are respectively subtypes of the first and second elements of [number, string] . However, after examining real world usage of tuples, we noticed that most situations in which this was permitted was typically undesirable.

In TypeScript 2.7, tuples of different arities are no longer assignable to each other. Thanks to a pull request from Kiara Grouwstra , tuple types now encode their arity into the type of their respective length property. This is accomplished by leveraging numeric literal types, which now allow tuples to be distinct from tuples of different arities.

Conceptually, you might consider the type [number, string] to be equivalent to the following declaration of NumStrTuple :

Note that this is a breaking change for some code. If you need to resort to the original behavior in which tuples only enforce a minimum length, you can use a similar declaration that does not explicitly define a length property, falling back to number .

Note that this does not imply tuples represent immutable arrays, but it is an implied convention.

Improved type inference for object literals

TypeScript 2.7 improves type inference for multiple object literals occurring in the same context. When multiple object literal types contribute to a union type, we now normalize the object literal types such that all properties are present in each constituent of the union type.

Previously type {} was inferred for obj and the second line subsequently caused an error because obj would appear to have no properties. That obviously wasn’t ideal.

Multiple object literal type inferences for the same type parameter are similarly collapsed into a single normalized union type:

Improved handling of structurally identical classes and instanceof expressions

TypeScript 2.7 improves the handling of structurally identical classes in union types and instanceof expressions:

  • Structurally identical, but distinct, class types are now preserved in union types (instead of eliminating all but one).
  • Union type subtype reduction only removes a class type if it is a subclass of and derives from another class type in the union.
  • Type checking of the instanceof operator is now based on whether the type of the left operand derives from the type indicated by the right operand (as opposed to a structural subtype check).

This means that union types and instanceof properly distinguish between structurally identical classes.

Type guards inferred from in operator

The in operator now acts as a narrowing expression for types.

For a n in x expression, where n is a string literal or string literal type and x is a union type, the “true” branch narrows to types which have an optional or required property n , and the “false” branch narrows to types which have an optional or missing property n .

Support for import d from "cjs" from CommonJS modules with --esModuleInterop

TypeScript 2.7 updates CommonJS/AMD/UMD module emit to synthesize namespace records based on the presence of an __esModule indicator under esModuleInterop . The change brings the generated output from TypeScript closer to that generated by Babel.

Previously CommonJS/AMD/UMD modules were treated in the same way as ES6 modules, resulting in a couple of problems. Namely:

  • TypeScript treats a namespace import (i.e. import * as foo from "foo" ) for a CommonJS/AMD/UMD module as equivalent to const foo = require("foo") .Things are simple here, but they don’t work out if the primary object being imported is a primitive or a class or a function. ECMAScript spec stipulates that a namespace record is a plain object, and that a namespace import ( foo in the example above) is not callable, though allowed by TypeScript
  • Similarly a default import (i.e. import d from "foo" ) for a CommonJS/AMD/UMD module as equivalent to const d = require("foo").default .Most of the CommonJS/AMD/UMD modules available today do not have a default export, making this import pattern practically unusable to import non-ES modules (i.e. CommonJS/AMD/UMD). For instance import fs from "fs" or import express from "express" are not allowed.

Under the new esModuleInterop these two issues should be addressed:

  • A namespace import (i.e. import * as foo from "foo" ) is now correctly flagged as uncallable. Calling it will result in an error.
  • Default imports to CommonJS/AMD/UMD are now allowed (e.g. import fs from "fs" ), and should work as expected.
Note: The new behavior is added under a flag to avoid unwarranted breaks to existing code bases. We highly recommend applying it both to new and existing projects. For existing projects, namespace imports ( import * as express from "express"; express(); ) will need to be converted to default imports ( import express from "express"; express(); ).

With esModuleInterop two new helpers are generated __importStar and __importDefault for import * and import default respectively. For instance input like:

Will generate:

Numeric separators

TypeScript 2.7 brings support for ES Numeric Separators . Numeric literals can now be separated into segments using _ .

Cleaner output in --watch mode

TypeScript’s --watch mode now clears the screen after a re-compilation is requested.

Prettier --pretty output

TypeScript’s pretty flag can make error messages easier to read and manage. pretty now uses colors for file names, diagnostic codes, and line numbers. File names and positions are now also formatted to allow navigation in common terminals (e.g. Visual Studio Code terminal).

Nightly Builds

How to use a nightly build of TypeScript

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Strict Property Initialization in TypeScript

TypeScript 2.7 introduced a new compiler option for strict property initialization checks in classes. If the --strictPropertyInitialization flag is enabled, the type checker verifies that each instance property declared in a class either

  • has a type that includes undefined ,
  • has an explicit initializer , or
  • is definitely assigned to in the constructor.

The --strictPropertyInitialization option is part of the family of compiler options that is enabled automatically when the --strict flag is set. As with all the other strict compiler options, you can set --strict to true and selectively opt out of strict property initialization checks by setting --strictPropertyInitialization to false .

Note that the --strictNullChecks flag must be set (either directly or indirectly via --strict ) in order for --strictPropertyInitialization to have any effect.

Alright, let's see strict property initialization checks in action. Without the --strictPropertyInitialization flag enabled, the following code type-checks just fine, but produces a TypeError at runtime:

The reason for the runtime error is that the username property holds the value undefined because there's no assignment to that property. Therefore, the call to the toLowerCase() method fails.

If we enable --strictPropertyInitialization , the type checker raises an error:

Let's look at four different ways we can properly type our User class to make the type error go away.

# Solution #1: Allowing undefined

One way to make the type error go away is to give the username property a type that includes undefined :

Now, it's perfectly valid for the username property to hold the value undefined . Whenever we want to use the username property as a string, though, we first have to make sure that it actually holds a string and not the value undefined , e.g. using typeof :

Alternatively, we can use optional chaining (the ?. operator) to only call the toLowerCase() method if the username property holds a non-nullish value. We can combine that with nullish coalescing (the ?? operator) to provide the fallback value:

# Solution #2: Explicit Property Initializer

Another way to make the type error go away is to add an explicit initializer to the username property. This way, the property holds a string value right away and is not observably undefined :

# Solution #3: Assignment in the Constructor

Perhaps the most useful solution is to add a username parameter to the constructor, which is then assigned to the username property. This way, whenever an instance of the User class is constructed, the caller has to provide the username as an argument:

We could simplify the User class by removing the explicit assignment to the class field and adding the public modifier to the username constructor parameter:

Note that strict property initialization requires each property to be definitely assigned in all possible code paths in the constructor. The following (contrived) example is therefore not type-correct because in some cases, we leave the username property uninitialized:

# Solution #4: Definite Assignment Assertion

If a class property neither has an explicit initializer nor a type including undefined , the type checker requires that property to be initialized directly within the constructor; otherwise, strict property initialization checks will fail. This is problematic if you want to initialize a property within a helper method or have a dependency injection framework initialize it for you. In these cases, you have to add a definite assignment assertion ( ! ) to that property's declaration:

By adding a definite assignment assertion to the username property, we're telling the type checker that it can expect the username property to be initialized, even if it cannot detect that on its own. It is now our responsibility to make sure the property is definitely assigned to after the constructor returns, so we have to careful; otherwise, the username property can be observably undefined and we're back to the TypeError at runtime.

This article and 44 others are part of the TypeScript Evolution series. Have a look!

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Strict Property Initialization in TypeScript

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In later versions of TypeScript there’s the concept of a Strict Class Property Initializer, or Definite Assignment Assertion .

This feature all of a sudden produced type errors on previously working code, so let’s explore why it’s here and why you should be using it.

“Property … has no initializer and is not definitely assigned in the constructor.”

Sound familiar? Welcome to strict initialization and the requirement of definite assignment.

So, what it is? It’s a safer way to declare properties and how you expect the data to be made available in your application.

If you see this error, your property is simply being defined with no value. Or at least, no obvious value.

It’s most likely that you’ve seen this error in an Angular context, for example:

So how do we ‘fix it’? Well, there are a few ways. First, we can declare a ! after the property (using definite assignment):

What it means is: “TypeScript, I know there is no data here yet, but I’m confident it will be there soon, just trust me!”.

This is the preferred option, and I’d encourage you to stick with it.

Alternatively, you can supply a default value, and then it’s not needed (but this involves writing and importing code that doesn’t really do anything, I’m not a fan but it works):

Part of the error we see is “not defined in the constructor”, so by also doing that we can ‘fix’ it (but don’t do this):

Why don’t do this? Well, we’ve removed ngOnInit which is a great practice and you should be using it. Angular knows about ngOnInit , whereas JavaScript knows about the constructor . Angular has no control, therefore we should move to a lifecycle hook.

💥 Read my deep-dive on the difference between NgOnInit and the constructor .

You can (at your own peril) disable this feature in your tsconfig.json file as well, and possibly need to disable other strict TypeScript compiler options:

In a non-Angular context, it’s likely you’ll want to structure your data, properties and initializations better to avoid introducing bugs and other errors.

👋 Hello TypeScript dev! If you enjoyed this post, you’ll love my advanced TypeScript courses that I’ve built over years of experience and hard work.

Using the ! definite assignment in Angular is a safe and acceptable solution to writing components, services, directives and more - as demonstrated with OnInit , we can be sure we’re adopting the correct practice.

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Using The "Definite Assignment Assertion" To Define Required Input Bindings In Angular 7.1.1

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The art of simplicity

Search this blog, typescript 2.7–definite assignment assertions.

While looking through some TypeScript 2.7 samples, I noticed the following syntax:

  • has a type that includes undefined ,
  • has an explicit initializer
  • is initialized within the constructor

This is problematic if you want to initialize a property within a helper method or have a dependency injection framework do it for you. To solve this situation you can use a definite assignment assertion using the exclamation mark: (!) . Now the type checker will no longer complain.

Remark: By using a definite assignment assertion it becomes your responsibility to make sure the property gets set.

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How to dynamically assign properties to an object in TypeScript

angular add definite assignment assertion to property

Editor’s note: This article was updated on 6 October 2023, introducing solutions like type assertions and the Partial utility type to address the TypeScript error highlighted.

How To Dynamically Assign Properties To An Object In TypeScript

JavaScript is a dynamically typed language, meaning that a variable’s type is determined at runtime and by what it holds at the time of execution. This makes it flexible but also unreliable and prone to errors because a variable’s value might be unexpected.

TypeScript, on the other hand, is a statically typed version of JavaScript — unlike JavaScript, where a variable can change types randomly, TypeScript defines the type of a variable at its declaration or initialization.

Dynamic property assignment is the ability to add properties to an object only when they are needed. This can occur when an object has certain properties set in different parts of our code that are often conditional.

In this article, we will explore some ways to enjoy the dynamic benefits of JavaScript alongside the security of TypeScript’s typing in dynamic property assignment.

Consider the following example of TypeScript code:

This seemingly harmless piece of code throws a TypeScript error when dynamically assigning name to the organization object:

An Error Is Thrown When Dynamically Assigning A Property To An Object

See this example in the TypeScript Playground .

The source of confusion, perhaps rightly justified if you’re a TypeScript beginner, is: how could something that seems so simple be such a problem in TypeScript?

The TL;DR of it all is that if you can’t define the variable type at declaration time, you can use the Record utility type or an object index signature to solve this. But in this article, we’ll go through the problem itself and work toward a solution that should work in most cases.

The problem with dynamically assigning properties to objects

Generally speaking, TypeScript determines the type of a variable when it is declared. This determined type stays the same throughout your application. There are exceptions to this rule, such as when considering type narrowing or working with the any type, but otherwise, this is a general rule to remember.

In the earlier example, the organization object is declared as follows:

There is no explicit type assigned to this variable, so TypeScript infers a type of organization based on the declaration to be {} , i.e., the literal empty object.

If you add a type alias, you can explore the type of organization :

Exploring The Literal Object Type

See this in the TypeScript Playground .

When you then try to reference the name prop on this empty object literal:

You receive the following error:

There are many ways to solve the TypeScript error here. Let’s consider the following:

Solution 1: Explicitly type the object at declaration time

This is the easiest solution to reason through. At the time you declare the object, go ahead and type it, and assign all the relevant values:

This eliminates any surprises. You’re clearly stating what this object type is and rightly declaring all relevant properties when you create the object.

However, this is not always feasible if the object properties must be added dynamically, which is why we’re here.

Solution 2: Use an object index signature

Occasionally, the properties of the object truly need to be added at a time after they’ve been declared. In this case, you can use the object index signature, as follows:

When the organization variable is declared, you can explicitly type it to the following: {[key: string] : string} .

You might be used to object types having fixed property types:

However, you can also substitute name for a “variable type.” For example, if you want to define any string property on obj :

Note that the syntax is similar to how you’d use a variable object property in standard JavaScript:

The TypeScript equivalent is called an object index signature.

angular add definite assignment assertion to property

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angular add definite assignment assertion to property

Moreover, note that you could type key with other primitives:

Solution 3: Use the Record Utility Type

The Record utility type allows you to constrict an object type whose properties are Keys and property values are Type . It has the following signature: Record<Keys, Type> .

In our example, Keys represents string and Type . The solution here is shown below:

Instead of using a type alias, you can also inline the type:

Using The Record Utility Type

Solution 4: Use the Map data type

A Map object is a fundamentally different data structure from an object , but for completeness, you could eliminate this problem if you were using Map .

Consider the starting example rewritten to use a Map object:

With Map objects, you’ll have no errors when dynamically assigning properties to the object:

Dark Background Typescript Playground Showing Map Object No Errors

This seems like a great solution at first, but the caveat is your Map object is weakly typed. You can access a nonexisting property and get no warnings at all:

See the TypeScript Playground .

This is unlike the standard object. By default, the initialized Map has the key and value types as any — i.e., new () => Map<any, any> . Consequently, the return type of the s variable will be any :

Dark Background Typescript Playground Showing Constant S With Type Any Indicated By Red Arrow

When using Map , at the very least, I strongly suggest passing some type information upon creation. For example:

s will still be undefined, but you won’t be surprised by its code usage. You’ll now receive the appropriate type for it:

Dark Background Typescript Playground Showing Properly Typed Map Value With Const S With Type String Undefined Indicated By Red Arrow

If you truly don’t know what the keys of the Map will be, you can go ahead and represent this at the type level:

And if you’re not sure what the keys or values are, be safe by representing this at the type level:

Solution 5: Consider an optional object property

This solution won’t always be possible, but if you know the name of the property to be dynamically assigned, you can optionally provide this when initializing the object as shown below:

If you don’t like the idea of using optional properties, you can be more explicit with your typing as shown below:

Solution 6: Leveraging type assertions

TypeScript type assertion is a mechanism that tells the compiler the variable’s type and overrides what it infers from the declaration or assignment. With this, we are telling the compiler to trust our understanding of the type because there will be no type verification.

We can perform a type assertion by either using the <> brackets or the as keyword. This is particularly helpful with the dynamic property assignment because it allows the properties we want for our object to be dynamically set because TypeScript won’t enforce them.

Let’s take a look at applying type assertions to our problem case:

Dark Background Typescript Playground Showing Dynamic Property Assignment Using Type Assertion

Note that with type assertions, the compiler is trusting that we will enforce the type we have asserted. This means if we don’t, for example, set a value for organization.name , it will throw an error at runtime that we will have to handle ourselves.

Solution 7: Use the Partial utility type

TypeScript provides several utility types that can be used to manipulate types. Some of these utility types are Partial , Omit , Required , and Pick .

For dynamic property assignments, we will focus specifically on the Partial utility type. This takes a defined type and makes all its properties optional. Thus, we can initialize our object with any combination of its properties, from none to all, as each one is optional:

In our example with the Partial utility type, we defined our organization object as the type partial Org , which means we can choose not to set a phoneNumber property:

Dark Background Typescript Playground Showing Dynamic Property Assignment Using Utility Type Partial

Grouping and comparing the options for adding properties in TypeScript

In this article, we explored the different options for setting properties dynamically in TypeScript. These options can be grouped together by their similarities.

Index/Key signatures

This group of options allows you to define the type of keys allowed without limiting what possible keys can exist. The options in this group include:

  • Using an object index signature
  • Using the Record utility type
  • Using the Map data type (with key/value typing)

With these, we can define that our object will take string indexes and decide what types to support as values, like String , Number , Boolean , or Any :

See in TypeScript Playground .

Pro: The main benefit of these methods is the ability to dynamically add properties to an object while still setting expectations for the potential types of keys and values.

Con: The main disadvantage of this way of defining objects is that you can’t predict what keys our objects will have and so some references may or may not be defined. An additional disadvantage is that if we decide to define our key signature with type Any , then the object becomes even more unpredictable.

Conditional/Optional properties

This set of object assignment methods shares a common feature: the definition of optional properties. This means that the range of possible properties are known but some may or may not be set. The options in this group include:

  • Using optional object properties
  • Using the Partial utility type
  • Using type assertions

See this example in the TypeScript Playground , or in the code block below:

Note: While these options mean that the possible keys are known and may not be set, TypeScript’s compiler won’t validate undefined states when using type assertions. This can lead to unhandled exceptions during runtime. For example, with optional properties and the Partial utility type, name has type string or undefined . Meanwhile, with type assertions, name has type string .

Pro: The advantage of this group of options is that all possible object keys and values are known.

Con: The disadvantage is that while the possible keys are known, we don’t know if those keys have been set and will have to handle the possibility that they are undefined.

Apart from primitives, the most common types you’ll have to deal with are likely object types. In cases where you need to build an object dynamically, take advantage of the Record utility type or use the object index signature to define the allowed properties on the object.

If you’d like to read more on this subject, feel free to check out my cheatsheet on the seven most-asked TypeScript questions on Stack Overflow, or tweet me any questions . Cheers!

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3 Replies to "How to dynamically assign properties to an object in TypeScript"

I know this is explicitly for TypeScript, and I think type declarations should always be first. But in general, you can also use a Map object. If it’s really meant to by dynamic, might as well utilize the power of Map.

Great suggestion (updated the article). It’s worth mentioning the weak typing you get by default i.e., with respect to Typescript.

Hi, thanks for your valuable article please consider ‘keyof type’ in TypeScript, and add this useful solution if you are happy have nice time

Leave a Reply Cancel reply

Property '...' has no initializer and is not definitely assigned in the constructor error fix in Angular

Fixing Property ‘…’ has no initializer and is not definitely assigned in the constructor error

If you are using latest version of Angular application, you might have encountered Property '...' has no initializer and is not definitely assigned in the constructor error, when you declare a variable inside a class or component.

Consider below example.

It’s because of Strict Class Initialization flag introduced in TypeScript 2.7 version.

So If the --strictPropertyInitialization typescript flag is enabled, the compiler checks if each property declared inside a class or angular component

  • has an undefined type
  • has an explicit initializer
  • or checks if it’s assigned in the constructor or not.

If not it throws Property '...' has no initializer and is not definitely assigned in the constructor error.

The flag is enabled by default if --strict flag in enabled in typescript compiler options.

To fix Property '...' has no initializer and is not definitely assigned in the constructor error in Angular applications use the below methods.

  • Disable strictPropertyInitialization flag
  • Adding undefined type to the property
  • Add definite assignment assertion to property
  • Add initializer to property
  • Assignment in the Constructor

Solution 1: Disable strictPropertyInitialization flag

The simple way to fix this error in Angular applications is to disable --strictPropertyInitialization flag in typescript compiler options in tsconfig.json file.

If not there other ways we can by pass this error.

Solution 2: Adding undefined type to the property

It’s ok to have an undefined property.

So while declaring variable add undefined type to the property.

But while using employees property we have to check for undefined value to avoid errors in run time.

Solution 3: Add definite assignment assertion to property

If you know that we will assign the property in later point in time.

It’s better to add definite assignment assertion to the property. i.e., employees.

To add the definite assignment assertion we have to add Exclamation mark(!) symbol after the variable name.

Be careful, we are just bypassing the compiler error, so it’s our responsibility to make sure the property is definitely assigned before using it. otherwise we will get TypeError at run time.

Solution 4: Add initializer to property

Another way to make this type error go away is to add an explicit initializer to the property.

Solution 5: Assignment in the Constructor

Otherwise, we can assign some default value to the property in the constructor.

Arunkumar Gudelli

Definite Assignment Assertion

Mike North

Check out a free preview of the full Intermediate TypeScript, v2 course

The "Definite Assignment Assertion" Lesson is part of the full, Intermediate TypeScript, v2 course featured in this preview video. Here's what you'd learn in this lesson:

Mike discusses the definite assignment operator and explains how to use the exclamation mark (!) to assert that a variable will be assigned a value before it is used. A question about using the declare keyword instead of the exclamation mark is also covered in this segment.

Transcript from the "Definite Assignment Assertion" Lesson

[00:00:00] >> The next thing we're gonna talk about is definite assignment assertion. It also involves an exclamation mark and to do this part, this exercise, we're going to need to go into our tsconfig for this notes project and we're gonna turn on strictPropertyInitialization and set it to true. You should see some errors pop up on the left, and we do, perfect.

[00:00:26] So make sure you turn that on, we're gonna talk about what that is, and how we deal with it. So it's strictPropertyInitialization. So I have a class here and it's called ThingWithAsyncSetup and I've done something that's a little odd here, I have something async inside of a constructor and some might argue I'm already doing something a bit weird.

[00:00:53] But I might pushback against that and say, well look, I'm kinda just kicking off the promise. I'm doing nothing in the constructor that depends on the promise being resolved, and it looks like I have some state I can have in place where I can from the outside observe whether whatever this setup promise does I can observe that it is successfully, has successfully completed, right?

[00:01:21] The dot then will be called and then I'll see this flip. And I have here isSetup as a class field set to a boolean but I'm getting an error, and the error is that is set up has no initializer, and it's not definitely assigned in the constructor. So if I were to take this one line here and move it up into the constructor, the error goes away, and that's because TypeScript analyzes the constructor, it looks at all the class fields that don't have an initializer which would look like this, right?

[00:01:54] I mean, you might argue in this scenario, I should do this, but this is a good example here. The problem here is Typescript saying, I don't know, it doesn't seem like, at least in the general case isSetup, like by the time we return an instance, is that actually gonna be a Boolean?

[00:02:16] It seems like it might be undefined. And why do we think that it's saying that? Does anyone have any hypotheses? Any guesses? I mean, I'm setting it in here. >> Is that run after the constructor? >> Typescript sees a callback here, and it's like, callbacks. I don't know when promise invokes this callback, I don't know if, by the time this promise function, the constructor for promise, by the time it returns, how do I know that this callback has been completed?

[00:02:58] It doesn't know effectively that once we advance beyond line 49, whether isSetup will be set, and in the general case, that is a perfectly valid thing to be concerned about, right? You could get an instance of thing with AsyncSetup, and it claims to have a boolean value for setup, but it could be undefined, except this callback here, we would call this like the promise executor, right?

[00:03:29] It's the thing you use to, it's the function that represents sort of the work or the beginning of the work that could be async, right? This is invoked synchronously within the promise constructor. So I know that by the time we assign the promise to setupPromise, this function's already done, it has been completed.

[00:03:55] I know, definitely, that isSetup, this line of code on line 50, that will have been run, I know for a fact. Now, I also know that TypeScript, there's no way to articulate that in TypeScript and to say, this is a synchronous function, you should assume that this is completed before, like chill out TypeScript, this has been run, there's no way to say that.

[00:04:19] But I can say, look, I know about this particular callback, and I'm gonna say, isSetup is definitely going to have been assigned by the time the constructor completes. Now what I've given you here, it's actually a very, this is a true example, and it does actually guarantee that by the time you get an instance of ThingWithAsyncSetup, it's gonna have a boolean value there, that is true.

[00:04:53] Sometimes you might know something else, like you might know that look, this object is gonna be created, but no one's gonna be working with it or using it until this thing is done to it. Maybe you're initializing objects and you're collecting them, and then after they're already initialized, you set some stuff up on them, and you want downstream consumers to not have to deal with the possibility that a bunch of things could be undefined.

[00:05:23] And so you might blur the lines here and say, okay, technically TypeScript, you're correct, these fields might not have been initialized, but I'll take care of initializing them myself from the outside. And the thing I wanna be sure of is that whoever is working with isSetup, they don't have to deal with this type here, they don't have to deal with the truly honest state of where it is after instantiation.

[00:05:51] You might wanna say, look, right after I create it, I'm doing this thing to it, and I'm always doing that, and I'll take responsibility for making sure that that's true and that no one gets a hold of these things before. So an example of where you might see this is like framework level code, where objects are created, not by invoking the constructor, but by some other process like they're created as a side effect of doing something.

[00:06:17] And then you can be assured that, whenever we go through this pathway where these things are created. I'm grabbing that instance and I'm doing a bunch of stuff to it, and then when I hand it off, I want it to be represented as if the stuff was always there.

[00:06:35] You're taking on risk there, but sometimes it's a valid pattern. Sometimes it's preferable over modeling an interface, or a class that represents, the partially set up thing, and then you have another class that you have to deserialize and serialize into so you get the fully set up thing, it can get complicated if you try to stick very rigidly to the rules.

[00:07:00] Those are two use cases for definitely assignment operator. >> Is it a good practice to use the non-knowledge assertion operator in production code? >> Non-null assertion operator. >> The exclamation point. >> The exclamation point, so what I said before, and I'll stick to it here. I use this all the time in my tests.

[00:07:23] I try to not use it in applications, in libraries, in any code that's actually running and the reason is the mode of failure is usually unacceptable to me, it'll just throw your codes interrupted, and it'll fail however, it's gonna fail. And I usually like to, like I would say, if you want this equivalent thing in production, what you should really do is this.

[00:08:03] Do this. And then write something explicit, like how to handle, the case where it's not there, just use a type card. In tests, this will get very messy cuz I might, on every single line, I get a big JSON object like back that has tons of properties and things on it.

[00:08:27] My test would be three quarters, like by line count. Type guards and safely finding if things are there and throwing very specific errors, it's not just that the outer object wasn't there, but the inner object too and the inner inner inner object. And that's where using this non-null assertion operator, it's valuable cuz all I want is to throw the, as I try to probe into this object, drill into this object, and this is a very concise way of doing that, it's easier to maintain, your test remains a lot more readable, but I don't use this.

[00:09:10] I'm not gonna speak in absolutes here, it would have to be exceptional case for me to use this in production code, not even a production code, any code that's outside of the testing sphere. Cuz you can always do this and this is much more graceful error handling, this is okay when throwing is the desired outcome, that is an absolute statement I'll make because that's exactly what I'm near it'll throw >> Is there a difference between [COUGH] using the exclamation point on 53 and using the declare keyword?

[00:09:55] >> Like this? That's interesting. I wouldn't necessarily do this. I'm actually surprised that this works. I use declare mostly in ambient type information, or when I'm making a mod, like a type declaration, which we'll see later, where you can do that, you can augment types in other modules or in the global scope, but I generally don't use this.

[00:10:30] It does appear to work, it appears to work. I would be kind of, if I saw this in a poll request, I would comment on it and say, you're asking people to understand that you can do this and there's an equally effective way to do this and it's much more common and people will know it when they come across it.

[00:10:53] But that's interesting, you taught me something here. I guess we're saying what we're doing there with declare, if I had to guess, I don't know this for a fact, but I'll explain how my mental model is processing that. When we say declare, we might be stating that, this is a boolean, it's of type boolean, and it's sort of a different thing then checking for definite assignment, right?

[00:11:24] You're really, yeah, that's the best way I can articulate it. I'd recommend using this, I don't think I've ever seen the use of declare before field unless it's in a declaration file, like a DTS file. >> I've ended up having to use it like in a number of projects where I'm injecting, setting that to a bad service, using that decorator.

[00:11:50] And it's probably cause like my TS config wasn't set up right or something, but it would complain about the exclamation point there but it wouldn't complain about the declares. >> I see like a collision of syntax. Yeah, I would use declare there given that it works, but man, is that uncommon and please file a bug report because I think that's odd.

[00:12:13] It's odd to the point where I'm not sure it will continue to work that way, it's that unusual. It'll work in declaration files, all of definitelytyped like that's declared everywhere but in this space, I've never seen that, interesting. I believe you though, there are all sorts of typescript syntax things that sort of collide with other things like this, this is a when you mix this with JSX, it's quite interesting here, this is why we don't write arrays like, This is popular for a long time, but this makes it complicated in JSX, where the parsers get overwhelmed with the same kind of syntax, meaning different things in different spaces.

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Late binding and TypeScript definite assignment assertions

If you take a close look at the code for the previous UserSettings example component, you may notice a weird exclamation mark:

Now, what is it and why did we need it?

As the official TypeScript documentation ( https://www.typescriptlang.org/docs/handbook/release-notes/typescript-2-7.html#definite-assignment-assertions ) states, definite assignment assertions can be used to tell the compiler that variables will indeed be assigned, even if TypeScript can't determine this on its own .

The definite assignment assertion is used through the ! character, and is defined after a property or variable declaration.

If we consider ...

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angular add definite assignment assertion to property

Property ‘…’ has no initializer and is not definitely assigned in the constructor in Angular

If you’re developing an Angular application using the latest version of TypeScript, you might encounter a compilation error: “Property ‘…’ has no initializer and is not definitely assigned in the constructor.” This error is due to the strict class initialization introduced in TypeScript 2.7.

This tutorial will guide you through various solutions provided by the community to resolve this issue.

Solution 1: Update tsconfig.json

Open your tsconfig.json file and set the strictPropertyInitialization flag to false under the compilerOptions . This disables strict property initialization and resolves the compilation error.

Solution 2: Change TypeScript version in VSCode

If you are using Visual Studio Code, you can change the TypeScript version that the editor uses. This can be done by configuring the TypeScript version in your editor settings.

Solution 3: Initialize Variables in the Constructor

Ensure that all class properties are initialized either when declared or in the constructor. For example:

Solution 4: Use Definite Assignment Assertion

If you are sure that a property will be assigned a value at runtime, you can use the definite assignment assertion ( ! ):

Solution 5: Adjust Input Bindings

When dealing with @Input properties, add the definite assignment assertion or use optional properties if the variable might be undefined:

Solution 6: Update tsconfig.json Strict Options

If the compilation error is related to strict options in tsconfig.json , adjust the following settings:

Solution 7: Assign Default Values

Assign default values to properties in the constructor to ensure they are initialized:

Choose the solution that best fits your project’s requirements. Whether it’s adjusting compiler options, initializing variables, or using definite assignment assertions, these approaches will help you overcome the “Property ‘…’ has no initializer” error in Angular applications.

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Understanding the "Property '...' has no initializer" Error in Angular and TypeScript

angular typescript

Understanding the Error:

This error arises in Angular applications built with TypeScript when you declare a property in a class but neglect to:

  • Assign it a value directly during declaration: class MyClass { myProperty : string ; // Error: No initializer }
  • Initialize it within the constructor: class MyClass { myProperty : string ; constructor ( ) { this .myProperty = "" ; // Initialization in constructor } }

Why Does This Error Happen?

TypeScript, by default, enforces strict class initialization. This ensures that all properties in a class have a definite value before they're used. This prevents potential runtime errors due to uninitialized variables.

Resolving the Error:

There are three primary approaches to address this error:

Initialize the Property During Declaration:

  • Assign a default value: class MyClass { myProperty : string = "" ; // Default value (empty string) }
  • Use undefined (if the property might initially be undefined): class MyClass { myProperty : string | undefined ; // Can be undefined }

Initialize the Property in the Constructor:

  • Explicitly assign a value within the constructor: class MyClass { myProperty : string ; constructor ( ) { this .myProperty = "" ; // Initialization } }

Definite Assignment Assertion (Not Recommended):

Choosing the Right Approach:

  • Default Values: Ideal for properties that have a sensible initial state.
  • Initialization in Constructor: Preferred for properties that rely on external data or complex logic.
  • Definite Assignment Assertion: Use sparingly, as it bypasses type safety checks and can lead to runtime issues if the property remains unassigned.

Additional Considerations:

  • Optional Properties: If a property might legitimately be undefined , make it optional using a question mark ( ? ): class MyClass { myOptionalProperty?: string ; }
  • Strict Null Checks: TypeScript's strict null checks (enabled by strictNullChecks in tsconfig.json ) can also contribute to this error. Ensure your properties are typed appropriately to avoid these checks.

By following these guidelines, you can effectively address the "Property '...' has no initializer" error in your Angular projects, maintaining type safety and preventing potential bugs.

Scenario 1: Missing Initialization

This code will trigger the error:

Solution 1: Default Value

Solution 2: Initialization in Constructor

Scenario 2: Optional Property

If a property might be undefined, use a question mark:

Scenario 3: Definite Assignment Assertion (Use with Caution!)

Remember, the definite assignment assertion ( ! ) should be used sparingly as it bypasses type safety checks. It's better to initialize the property properly for predictable behavior.

Default Values:

  • You can use different default values depending on the property type: class User { name : string = "" ; age: number = 0 ; isActive: boolean = false ; // Default values for different types }

Initialization in Constructor (Variations):

  • You can fetch data from services or use dependency injection within the constructor to initialize properties: class ProductService { constructor ( private http: HttpClient ) {} getProduct ( id: number ) { // ... fetch product data using http return productData; } } class ProductComponent { product : Product; constructor ( private productService: ProductService ) { this .product = this .productService.getProduct( 123 ); // Initialize with fetched data } }

Optional Properties:

  • You can make the property optional using a question mark ( ? ) along with a type that allows undefined : class User { name : string ; address?: string ; // Optional address property }

Nullish Coalescing Operator (Conditional Default Value):

  • In situations where you want a default value only if the property is null or undefined , use the nullish coalescing operator ( ?? ): class Settings { theme : string | undefined ; get defaultTheme () { return this .theme ?? 'light' ; // Use 'light' if theme is null or undefined } }

Remember, the definite assignment assertion ( ! ) should be a last resort, as it weakens type safety.

By understanding these variations and using them appropriately, you can effectively manage property initialization in your Angular and TypeScript applications.

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COMMENTS

  1. What's the difference between definite assignment assertion and ambient

    name!: string; This says to the compiler: "There is a property called name with a type of string | undefined.It starts with a value of undefined.But every time I get or set that property, I want to treat it as type string.". Using this form it's clear to anyone reading the code that name is undefined at first, but is treated like a string anyway. That means it must be set in this file ...

  2. Using The "Definite Assignment Assertion" To Define Required Input

    Ben Nadel looks at the "Definite Assignment Assertion" introduced in TypeScript 2.7. This assertion makes it possible to declare input bindings as "required" on an Angular Directive without having to fully initialize the internal class property. This makes the code easier to work with; and, expresses better intent in the code.

  3. TypeScript definite assignment assertions

    Fixing, and definite assignment assertions. The correct fix is probably to assign this.wordsPerMinute = wordsPerMinute in the constructor - but in some cases, you may be doing something funky where the dependency will be resolved in a way the compiler is unable to determine. When you need to allow a property with no definite assignment, you ...

  4. TypeScript: Documentation

    In our example, we knew that all uses of x would be initialized so it makes more sense to use definite assignment assertions than non-null assertions.. Fixed Length Tuples. In TypeScript 2.6 and earlier, [number, string, string] was considered a subtype of [number, string].This was motivated by TypeScript's structural nature; the first and second elements of a [number, string, string] are ...

  5. Strict Property Initialization in TypeScript

    #Solution #4: Definite Assignment Assertion. If a class property neither has an explicit initializer nor a type including undefined, the type checker requires that property to be initialized directly within the constructor; otherwise, strict property initialization checks will fail. This is problematic if you want to initialize a property ...

  6. Strict Property Initialization in TypeScript

    Trusted by 82,951 students. In later versions of TypeScript there's the concept of a Strict Class Property Initializer, or Definite Assignment Assertion. This feature all of a sudden produced type errors on previously working code, so let's explore why it's here and why you should be using it. "Property … has no initializer and is not ...

  7. Using The "Definite Assignment Assertion" To Define Required Input

    Using The "Definite Assignment Assertion" To Define Required Input Bindings In Angular 7.1.1 Loading files... npm Run Scripts: npm run build — Compiles the .ts file into bundles.; npm run watch — Compiles the .ts file into bundles and then watches files for changes. — Compiles the .ts file into bundles and then watches files for changes.

  8. Using The "Definite Assignment Assertion" To Define Required Input

    // "Definite Assignment Assertion" (!) to tell TypeScript that we know this value // will be defined in some way by the time we use it; and, that TypeScript should // not worry about the value until then. public user!: User; // I initialize the badge component. constructor() {// We've used the "Definite Assignment Assertion" to tell TypeScript ...

  9. Strict Property Initialization in TypeScript

    Mar 7, 2023. In later versions of TypeScript there's the concept of a Strict Class Property Initializer, or Definite Assignment Assertion. This feature all of a sudden produced type errors on ...

  10. TypeScript 2.7-Definite Assignment Assertions

    This is problematic if you want to initialize a property within a helper method or have a dependency injection framework do it for you. To solve this situation you can use a definite assignment assertion using the exclamation mark: (!) . Now the type checker will no longer complain. ... But when I tried to run an Angular app a little bit later ...

  11. How to dynamically assign properties to an object in TypeScript

    Solution 1: Explicitly type the object at declaration time. This is the easiest solution to reason through. At the time you declare the object, go ahead and type it, and assign all the relevant values: type Org = {. name: string. } const organization: Org = {. name: "Logrocket" } See this in the TypeScript Playground.

  12. Property '...' has no initializer and is not definitely assigned in the

    Solution 3: Add definite assignment assertion to property. If you know that we will assign the property in later point in time. It's better to add definite assignment assertion to the property. i.e., employees. employees!: Employee[]; To add the definite assignment assertion we have to add Exclamation mark(!) symbol after the variable name.

  13. Definite Assignment Assertion

    Here's what you'd learn in this lesson: Mike discusses the definite assignment operator and explains how to use the exclamation mark (!) to assert that a variable will be assigned a value before it is used. A question about using the declare keyword instead of the exclamation mark is also covered in this segment. Get Unlimited Access Now. Preview.

  14. Late binding and TypeScript definite assignment assertions

    Late binding and TypeScript definite assignment assertions If you take a close look at the code for the previous UserSettings example component, you may notice a weird exclamation mark: @Inject("username") … - Selection from Learn TypeScript 3 by Building Web Applications [Book]

  15. Property '...' has no initializer and is not definitely assigned in the

    Solution 4: Use Definite Assignment Assertion. If you are sure that a property will be assigned a value at runtime, you can use the definite assignment assertion (!): makes!: any[]; Solution 5: Adjust Input Bindings. When dealing with @Input properties, add the definite assignment assertion or use optional properties if the variable might be ...

  16. StrictPropertyInitialization in TypeScript

    Use the definite assignment assertion typescript. The definite assignment assertion is a feature that typescript introduced in version 2.7. We can use it to tell the typescript compiler that we will take care of giving the property its initial value. We do that by placing ! after instance property ( also any variable declaration).

  17. How to create custom type in angular

    EDIT 12/28/21: Angular by default now has strict typing on by default. This means that there is a bit more work to do when declaring a class with a set of properties. Each property must be changed using one of these techniques: Add | undefined to the type; Add a default value such as = '' Use the definite assignment assertion operator (!

  18. Understanding the "Property '...' has no initializer" Error in Angular

    Initialize the Property in the Constructor: Explicitly assign a value within the constructor: class MyClass { myProperty: string; constructor { this.myProperty = ""; // Initialization} } Definite Assignment Assertion (Not Recommended): Use the non-null assertion operator (!

  19. angular

    Background. I have a template that looks like this (I'm using some component that uses this as the basis for a repeated item, it's the <p-pickList>, but the question is not specific about that component, just as an example). For background, let's say I have a type Foo and my component has a foos: Foo[], I'm feeding it to the <p-pickList> component under the [source] attribute and its doing the ...