ChatGPT解决这个技术问题 Extra ChatGPT

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

in my Angular app i have a component:

import { MakeService } from './../../services/make.service';
import { Component, OnInit } from '@angular/core';

@Component({
  selector: 'app-vehicle-form',
  templateUrl: './vehicle-form.component.html',
  styleUrls: ['./vehicle-form.component.css']
})
export class VehicleFormComponent implements OnInit {
  makes: any[];
  vehicle = {};

  constructor(private makeService: MakeService) { }

  ngOnInit() {
    this.makeService.getMakes().subscribe(makes => { this.makes = makes
      console.log("MAKES", this.makes);
    });
  }

  onMakeChange(){
    console.log("VEHICLE", this.vehicle);
  }
}

but in the "makes" property I have a mistake. I dont know what to do with it...

https://i.stack.imgur.com/QpJLJ.png


M
Martin Čuka

Just go to tsconfig.json and set

"strictPropertyInitialization": false

to get rid of the compilation error.

Otherwise you need to initialize all your variables which is a little bit annoying


Just make sure you add that after "strict": true else the transpiler seems to turn it on again (although VS seems to know it's off).
In this way you will disable the strict checking property initialization for all the project. It's better to add the ! postfix operator to the variable name, to just ignore this case, or initialize the variable inside the constructor.
You are suggesting to ignore potential problems that the compiler is telling about, this is not really safe. So downvote.
StrictPropertyInitialzer is a flag introduced in typescript 2.7. It's up to everybody to choose if you want to enable all these flags and strictly follow all rules or turn off some validation. It makes completely no sense to follow all the rules every time. If your code gets too verbose and downsides and are bigger than advantages you should definetly turn it off. I am not telling that this is the best practice in all cases but its definetly viable option in most cases...
As I mentioned in my answer you have two options either you disable validation OR initialize all variables. Its up to everybody to choose what will you benefit more from project to project.
C
Clashsoft

I think you are using the latest version of TypeScript. Please see the section "Strict Class Initialization" in the link.

There are two ways to fix this:

A. If you are using VSCode you need to change the TS version that the editor use.

B. Just initialize the array when you declare it

makes: any[] = [];

or inside the constructor:

constructor(private makeService: MakeService) { 
   // Initialization inside the constructor
   this.makes = [];
}

as the error says you need to initialize the variable to some value makes: any[] = [];
You must use 'Definite Assignment Assertion' to tell typescript that this variable will have a value at runtime as follows: makes!: any[];
HI @Sajeetharan, how to chnage vscode ts version
Initializing variables in the constructor is an Angular anti-pattern. Don't do it.
@KumaresanSd ctrl+shift+p and type typescript in the dropdown, then select "Select TypeScript Version"
a
anothernode

It is because TypeScript 2.7 includes a strict class checking where all the properties should be initialized in the constructor. A workaround is to add the ! as a postfix to the variable name:

makes!: any[];

The "!" syntax exists for those common-ish cases where you can't guarantee that the value will be defined immediately. It's an escape hatch, and shouldn't be relied on, as it can make your code less safe. A default value is usually preferred. Good to know it exists, though
@kingdaro is right. While this can generally be used, it can also lead to code that flat doesn't work. As an example, in the basic webapp generated by VS2017, change the assignment for the forecasts in fetchdata.components.ts by adding the '!' (public forecasts!: WeatherForecast[];) and it will cause it to completely error out
This is the best solution, since it is directly after the @Input() decorator (in the new angular), when reading any given component, it reads naturally and culls out any dev mistakes.
incidentally what is the difference from using ! or ? to handle nulls or undefined?
I needed to define a property that uses an Enum in a service on an app where I can't assume the user's choices, not even a default. It is defined in the first step and not read until the second, so this was very useful escape, good to know it exists, thanks very much.
S
Stephane

We may get the message Property has no initializer and is not definitely assigned in the constructor when adding some configuration in the tsconfig.json file so as to have an Angular project compiled in strict mode:

"compilerOptions": {
  "strict": true,
  "noImplicitAny": true,
  "noImplicitThis": true,
  "alwaysStrict": true,
  "strictNullChecks": true,
  "strictFunctionTypes": true,
  "strictPropertyInitialization": true,

Indeed the compiler then complains that a member variable is not defined before being used.

For an example of a member variable that is not defined at compile time, a member variable having an @Input directive:

@Input() userId: string;

We could silence the compiler by stating the variable may be optional:

@Input() userId?: string;

But then, we would have to deal with the case of the variable not being defined, and clutter the source code with some such statements:

if (this.userId) {
} else {
}

Instead, knowing the value of this member variable would be defined in time, that is, it would be defined before being used, we can tell the compiler not to worry about it not being defined.

The way to tell this to the compiler is to add the ! definite assignment assertion operator, as in:

@Input() userId!: string;

Now, the compiler understands that this variable, although not defined at compile time, shall be defined at run-time, and in time, before it is being used.

It is now up to the application to ensure this variable is defined before being used.

As an an added protection, we can assert the variable is being defined, before we use it.

We can assert the variable is defined, that is, the required input binding was actually provided by the calling context:

private assertInputsProvided(): void {
  if (!this.userId) {
    throw (new Error("The required input [userId] was not provided"));
  }
}

public ngOnInit(): void {
  // Ensure the input bindings are actually provided at run-time
  this.assertInputsProvided();
}

Knowing the variable was defined, the variable can now be used:

ngOnChanges() {
  this.userService.get(this.userId)
    .subscribe(user => {
      this.update(user.confirmedEmail);
    });
}

Note that the ngOnInit method is called after the input bindings attempt, this, even if no actual input was provided to the bindings.

Whereas the ngOnChanges method is called after the input bindings attempt, and only if there was actual input provided to the bindings.


this is the correct answer when using 'strict' mode
A lot of this makes sense to me apart from one thing... What would the assert give you? Would be fine for debugging, but not for using the app. Throwing errors that will not be caught is not a good idea. You should provide some fallback that will make things work no matter what.
This should be upvoted as the correct answer, as it solves the problem, rather than just turning of part of the linting.
Using "definite assignment assertion" in strict mode just to avoid the condition checks doesn't make great sense. You are undermining the whole logic of having a strict typed system. If avoiding those checks is more important to you, then you probably shouldn't use strict mode.
c
crispengari

Go to your tsconfig.json file and change the property:

 "noImplicitReturns": false

and then add

 "strictPropertyInitialization": false

under "compilerOptions" property.

Your tsconfig.json file should looks like:


{
      ...
      "compilerOptions": {
            ....
            "noImplicitReturns": false,
            ....
            "strictPropertyInitialization": false
      },
      "angularCompilerOptions": {
         ......
      }  
 }

Hope this will help !!

Good Luck


IMO, it is kinda going backward.
Firstly, thanks for the solution, it worked. Secondly, can anyone explain what just happened?
It worked , it is necessary when creating directives in angular.
A
Alexandre Annic

The error is legitimate and may prevent your app from crashing. You typed makes as an array but it can also be undefined.

You have 2 options (instead of disabling the typescript's reason for existing...):

1. In your case the best is to type makes as possibily undefined.

makes?: any[]
// or
makes: any[] | undefined

So the compiler will inform you whenever you try to access to makes that it could be undefined. Otherwise, if the // <-- Not ok lines below were executed before getMakes finished or if getMakes failed, your app would crash and a runtime error would be thrown. That's definitely not what you want.

makes[0] // <-- Not ok
makes.map(...) // <-- Not ok

if (makes) makes[0] // <-- Ok
makes?.[0] // <-- Ok
(makes ?? []).map(...) // <-- Ok

2. You can assume that it will never fail and that you will never try to access it before initialization by writing the code below (risky!). So the compiler won't take care about it.

makes!: any[] 

More specifically,

Your code design could be better. Defining a local and mutable variable is not a good practice. You should manage data storage inside your service:

firstly to be nullsafe,

secondly to be able to factorise a lot of code (including typing, loading state and errors)

finally to avoid mulitple and useless reftech.

The exemple below try to show this but I didn't tested it and it could be improved:

type Make = any // Type it

class MakeService {

  private readonly source = new BehaviorSubject<Make[] | undefined>(undefined);
  loading = false;

  private readonly getMakes = (): Observable<Make[]> => {
    /* ... your current implementation */
  };

  readonly getMakes2 = () => {
    if (this.source.value) {
      return this.source.asObservable();
    }
    return new Observable(_ => _.next()).pipe(
      tap(_ => {
        this.loading = true;
      }),
      mergeMap(this.getMakes),
      mergeMap(data => {
        this.source.next(data);
        return data;
      }),
      tap(_ => {
        this.loading = false;
      }),
      catchError((err: any) => {
        this.loading = false;
        return throwError(err);
      }),
    );
  };
}

@Component({
  selector: 'app-vehicle-form',
  template: `
    <div *ngIf="makeService.loading">Loading...</div>
    <div *ngFor="let vehicule of vehicules | async">
      {{vehicle.name}}
    </div>
  `,
  styleUrls: ['./vehicle-form.component.css']
})
export class VehicleFormComponent implements OnInit {
  constructor(public makeService: MakeService) {}

  readonly makes = this.makeService.getMakes2().pipe(
    tap(makes => console.log('MAKES', makes))
  );

  readonly vehicules = this.makes.pipe(
    map(make => make/* some transformation */),
    tap(vehicule => console.log('VEHICLE', vehicule)),
  );

  ngOnInit() {
    this.makes.subscribe(makes => {
      console.log('MAKES', makes);
    });
  }
}

This seems like the only correct solution. I'm horrified by the other answers.
I also consider that this error message is useful. So deactivate it in tsconfig.json is not the appropriate solution. And adding '!' can allows the compiler to let go, but it makes the code less sure, unless you're really sure of what is going on. A better solution seems to understand that the property can be undedinfed, and therefore the developer has to write code accordingly. This is the purpose of the answer to me.
@Rei Miyasaka, this "feature" was added later to typescript does that mean that all code that was written before is wrong ? No it does not. When you have your hello world application feel free to follow all the rules. But once you have huge project (which is a reason for language like typescript to start with) then you should decide how to write the code and all should follow it. I don't see ANYTHING wrong with disabling this particular option in tsconfig.json.
@MartinČuka Just because everyone was doing it doesn't mean it wasn't wrong or that it wasn't horrifying. TypeScript itself was brought about because JavaScript is kind of horrifying. And the larger the project, the more you should want to be strict with type rules.
option 2 adding a ! before : is more common.
S
Sumit

Update for 2021 :

there is property like "strictPropertyInitialization"

Just go to tsconfig.json and set

"strict": false

to get rid of the compilation error.

Otherwise you need to initialize all your variables which is a little bit annoying.

reason behind this error is :

typescript is a kind of more Secure lang as compare to javascript.

although this security is enhance by enabling strict feature .So every time when you initialize a variable typescript wants them to assign a value.


Horrible advice, throws out type safety so that you see fewer errors, rather than surrendering some convenience for safety.
k
kshetline

You either need to disable the --strictPropertyInitialization that Sajeetharan referred to, or do something like this to satisfy the initialization requirement:

makes: any[] = [];

B
Bronzdragon

You can also do the following if you really don't want to initialise it.

makes?: any[];

A
Astrophage

If you want to initialize an object based on an interface you can initialize it empty with following statement.

myObj: IMyObject = {} as IMyObject;

M
Md Wahid

Put a question (?) mark after makes variable.

  makes?: any[];
  vehicle = {};

  constructor(private makeService: MakeService) { }

It should now works. I'm using angular 12 and it works on my code.


c
codejockie

As of TypeScript 2.7.2, you are required to initialise a property in the constructor if it was not assigned to at the point of declaration.

If you are coming from Vue, you can try the following:

Add "strictPropertyInitialization": true to your tsconfig.json

If you are unhappy with disabling it you could also try this makes: any[] | undefined. Doing this requires that you access the properties with null check (?.) operator i.e. this.makes?.length

You could as well try makes!: any[];, this tells TS that the value will be assigned at runtime.


x
xgqfrms

there are many solutions

demo code

class Plant {
  name: string;
  // ❌ Property 'name' has no initializer and is not definitely assigned in the constructor.ts(2564)
}

solutions

add 'undefined' type

class Plant {
  name: string | undefined;
}

add definite assignment assertion

class Plant {
  name!: string;
}

declare with init value 👍

class Plant {
  name: string = '';
}

use the constructor with an init value 👍

class Plant {
  name: string;
  constructor() {
    this.name = '';
  }
}

use the constructor and init value by params 👍

class Plant {
  name: string;
  constructor(name: string) {
    this.name = name ?? '';
  }
}

shorthand of the 5 👍

class Plant {
  constructor(public name: string = name ?? '') {
    // 
  }
}

tsconfig.json

not recommended

{
  "compilerOptions": {
+   "strictPropertyInitialization": false,
-   "strictPropertyInitialization": true,
  }
}

refs

https://www.typescriptlang.org/docs/handbook/2/classes.html#--strictpropertyinitialization


T
TheMysterious

A batter approach would be to add the exclamation mark to the end of the variable for which you are sure that it shall not be undefined or null, for instance you are using an ElementRef which needs to be loaded from the template and can't be defined in the constructor, do something like below

class Component {
 ViewChild('idname') variable! : ElementRef;
}

s
samivic

if you don't want to change your tsconfig.json, you can define your class like this:

class Address{
  street: string = ''
}

or, you may proceed like this as well:

class Address{
  street!: string
}

by adding an exclamation mark "!" after your variable name, Typescript will be sure that this variable is not null or undefined.


P
Pinaki

Get this error at the time of adding Node in my Angular project -

TSError: ? Unable to compile TypeScript: (path)/base.api.ts:19:13 - error TS2564: Property 'apiRoot Path' has no initializer and is not definitely assigned in the constructor. private apiRootPath: string;

Solution -

Added "strictPropertyInitialization": false in 'compilerOptions' of tsconfig.json.

my package.json -

"dependencies": {
    ...
    "@angular/common": "~10.1.3",
    "@types/express": "^4.17.9",
    "express": "^4.17.1",
    ...
}

Ref URL - https://www.ryadel.com/en/ts2564-ts-property-has-no-initializer-typescript-error-fix-visual-studio-2017-vs2017/


M
M Komaei

in tsconfig.json file , inside "compilerOptions" add :

"strictPropertyInitialization": false,

https://i.stack.imgur.com/SUiVW.png


M
Masoud Bimmar

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

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.

"compilerOptions": {
  ///
  ,
  "strictPropertyInitialization":false
}

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.

employees: Employee[];

//Change to 

employees : Employee[] | undefined;

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[];

Solution 4: Add initializer to property

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

employees: Employee[] = [];

Solution 5: Assignment in the Constructor

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

employees: Employee[];

constructor() { 
    this.employees=[];
}

Best Solutions from here


J
Jai Govind Gupta

When you upgrade using typescript@2.9.2 , its compiler strict the rules follows for array type declare inside the component class constructor.

For fix this issue either change the code where are declared in the code or avoid to compiler to add property "strictPropertyInitialization": false in the "tsconfig.json" file and run again npm start .

Angular web and mobile Application Development you can go to www.jtechweb.in


A
Alexei - check Codidact

Another way to fix in the case when the variable must remain uninitialized (and it is dealt with at the run time) is to add undefined to the type (this is actually suggested by VC Code). Example:

@Input() availableData: HierarchyItem[] | undefined;
@Input() checkableSettings: CheckableSettings | undefined;

Depending on actual usage, this might lead to other issues, so I think the best approach is to initialize the properties whenever possible.


Worked for me when initializing the @ViewChild variable.. see for reference: stackblitz.com/edit/…
Writing @Input() availableData: HierarchyItem[] | undefined; is the same as writing @Input() availableData?: HierarchyItem[] ;)
T
Takesha Kaluarachchi

a new version of typescript has introduced strick class Initialization, that is means by all of the properties in your class you need to initialize in the constructor body, or by a property initializer. check it in typescript doccumntation to avoid this you can add (! or ?) with property

make!: any[] or make? : any[] 

otherwise, if you wish to remove strick class checking permanently in your project you can set strictPropertyInitialization": false in tsconfig.json file

"compilerOptions": { .... "noImplicitReturns": false, .... "strictPropertyInitialization": false },


What is the difference between '!' and '?' ?
@pouyada the ! is assertive assignment declaration saying that no matter what it is assigned even if typescript can't pick it up. ? is optional meaning it may or may not be there but your component should be prepared for either case.
A
Andy

Can't you just use a Definite Assignment Assertion? (See https://www.typescriptlang.org/docs/handbook/release-notes/typescript-2-7.html#definite-assignment-assertions)

i.e. declaring the property as makes!: any[]; The ! assures typescript that there definitely will be a value at runtime.

Sorry I haven't tried this in angular but it worked great for me when I was having the exact same problem in React.


Working with all versions since it is in the official documentation and I tried it. THANKS!!
O
Onur Dikmen

1) You can apply it like the code below. When you do this, the system will not give an error.

"Definite Assignment Assertion" (!) to tell TypeScript that we know this value

Detail info

@Injectable()
export class Contact {
  public name !:string;
  public address!: Address;
  public digital!: Digital[];
  public phone!: Phone[];
}

2) The second method is to create a constructor and define values here.

export class Contact {
  public name :string;
  public address: Address;
  public digital: Digital[];
  public phone: Phone[];

  constructor( _name :string,
     _address: Address,
     _digital: Digital[],
     _phone: Phone[])
  {
    this.name=_name;
    this.address=_address;
    this.digital=_digital;
    this.phone=_phone;
  }
}

3) The third choice will be to create a get property and assign a value as follows

  export class Contact {
      public name :string="";
      public address: Address=this._address;
    
      get _address(): Address {
        return new Address();
      }
     
    }

S
Sydney_dev

Add these two line on the tsconfig.json

"noImplicitReturns": true,
"strictPropertyInitialization": false,

and make sure strict is set to true


A
Abilogos

change the

fieldname?: any[]; 

to this:

fieldname?: any; 

k
keivan kashani

You can declare property in constructor like this:

export class Test {
constructor(myText:string) {
this.myText= myText;
} 

myText: string ;
}

m
maxisam

This has been discussed in Angular Github at https://github.com/angular/angular/issues/24571

I think this is what everyone will move to

quote from https://github.com/angular/angular/issues/24571#issuecomment-404606595

For angular components, use the following rules in deciding between:
a) adding initializer
b) make the field optional
c) leave the '!'

If the field is annotated with @input - Make the field optional b) or add an initializer a).
If the input is required for the component user - add an assertion in ngOnInit and apply c.
If the field is annotated @ViewChild, @ContentChild - Make the field optional b).
If the field is annotated with @ViewChildren or @ContentChildren - Add back '!' - c).
Fields that have an initializer, but it lives in ngOnInit. - Move the initializer to the constructor.
Fields that have an initializer, but it lives in ngOnInit and cannot be moved because it depends on other @input fields - Add back '!' - c).

A
Alexey Ryzhkov

In my case it works with different declaration according new typescript strict features:

@ViewChild(MatSort, {static: true}) sort!: MatSort;

If disable typescript new strict feature in tsonfig.json with

"compilerOptions": {
  ///
  ,
  "strictPropertyInitialization":false
}

the old Angular's guide code works well

@ViewChild(MatSort) sort: MatSort;

Here is 4 ways to solve the issue thanks to Arunkumar Gudelli (2022) https://www.angularjswiki.com/angular/property-has-no-initializer-and-is-not-definitely-assigned-in-the-constructor/


T
Tabish Zaman

JUST go to the tsconfig.ts and add "strictPropertyInitialization": false, into the compilerOptions Object .

if it doesn't solve yet, kindly re-open your Code Editor.

EXAMPLE:

"compilerOptions" : {
  "strictPropertyInitialization": false,
}

S
Suresh

Comment the //"strict": true line in tsconfig.json file.


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