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

Question

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...

Answer 1

Just go to tsconfig.json and set

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

to get rid of the compilation error.

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

Answer 2

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 = [];
}

Answer 3

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[];

Answer 4

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.

Answer 5

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

Answer 6

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's design could be improved. It is generally not considered best practice to define local and mutable variables. Instead, you should manage data storage within a service:

• Firstly to ensure null/undefined safety,
• Secondly to be able to factorise a lot of code (including typing, loading state and errors),
• Finally to avoid multiple and useless reftech across components.

Here is an example illustrating the concept I am suggesting. Please note that I have not tested it, and there is room for further improvement.

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);
    });
  }
}

Answer 7

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

1. add 'undefined' type

class Plant {
  name: string | undefined;
}

2. add definite assignment assertion

class Plant {
  name!: string;
}

3. declare with init value

class Plant {
  name: string = '';
}

4. use the constructor with an init value

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

5. use the constructor and init value by params

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

6. 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

Answer 8

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.

Answer 9

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

makes: any[] = [];

Answer 10

1. Go to tsconfig.json

2. add "strictPropertyInitialization": false, in "compilerOptions":

Answer 11

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

makes?: any[];

Answer 12

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

myObj: IMyObject = {} as IMyObject;

Answer 13

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.

Answer 14

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.

Answer 15

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.

Answer 16

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;
}

Answer 17

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

"strictPropertyInitialization": false,

Answer 18

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/

Answer 19

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

Answer 20

It is because typescript 2.7.2 included a strict class checking where all properties should be declared in constructor. So to work around that, just add an exclamation mark (!) like:

name!:string;

Answer 21

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.

Answer 22

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.

Answer 23

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 },

Answer 24

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();
      }
     
    }

Answer 25

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/

Answer 26

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

I think this recommendation from angular gives good advice:

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 a).
• Fields that have an initializer, but it lives in ngOnInit and cannot be moved because it depends on other @input fields - Add back '!' - c).

Answer 27

Add these two line on the tsconfig.json

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

and make sure strict is set to true

Answer 28

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,
}

Answer 29

change the

fieldname?: any[]; 

to this:

fieldname?: any; 

Answer 30

You can declare property in constructor like this:

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

myText: string ;
}

Answer 31

Follow 2 steps to solve this::

1. "strictPropertyInitialization": false entry in tsconfig.json
2. ctrl+c stop your server on Terminal(check for terminal, where it is running) and run it again with the command you use like ng serve These 2 steps, should solve your issue.. as it solved mine.. cheers.. ;)

Answer 32

Instead of turning off the strict mode, you can assign a initial value to the variable. For example:

1. makes: any[] = [null];

2. private year: number = 0;

Answer 33

A little late for the party. but the most voted answer includes changing the security level as a whole abandoning type safety, to work around this use the typing mechanism of the typescript itself

@ViewChild(MatPaginator) paginator: MatPaginator | undefined;

ngAfterViewInit() { 
  this.dataSource.paginator = this.paginator as MatPaginator;
}

I believe the angular material documentation should be updated, but until then I believe this is the best way to do it

Answer 34

You can also add @ts-ignore to silence the compiler only for this case:

//@ts-ignore
makes: any[];

Answer 35

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

Answer 36

Just declare your variable as type any

@Input() headerName: any;

Answer 37

Next to variables "?" You can fix it by putting it.

Example:

--------->id?:number --------->name?:string