如何编写将返回的函数：Observable<ModelTemp>

Question

How to write a function which will be return : Observable .

my service:

 modelTemp= [{ 
    name: 'abcdefghijk'
 }];   
 get(): Observable<ModelTemp> {
    return this.modelTemp;
 }

In service the method the variable is displayed red:

Returned expression type {name: string}[] is not assignable to type Observable less... (Ctrl+F1) Checks TypeScript called function parameters , return values , assigned expressions to be of correct type

How to fix this error?

my component:

 model: ModelTemp; 

    constructor(private getService: GetService) {}
    ngOnInit() {
      this.generate();
    }

    generate() {
      this.getService.get()
        .subscribe((modelData) => { 
          this.model = modelData;
        });
    }

my interface:

   export interface ModelTemp {
      name: string; 
   }

I do not know where I'm making a mistake. The model component should be declared differently (in this way): modelPage: ModelPage; or so: modelPage: ModelPage[];

In my the browser returns me an error:

this.getService.get(...).subscribe is not a function at TempComponent.push../src/app/temp/temp.component.ts.TempComponent.generate.

I want to use the subscribe method. Which will return the model (interface ModelTemp)

How to correct the code?

How to write a function which will be return : Observable .

Answer 1

use rxjs operators like of to return an Observable

import {of} from 'rxjs';

 get(): Observable<ModelTemp> {
  return of(this.modelTemp);
} 

Answer 2

You can try something like this:

modelTemp: BehaviorSubject<ModelTemp[]> = new BehaviorSubject([{ 
    name: 'abcdefghijk'
}]);

get(): BehaviorSubject <ModelTemp[]> {
    return this.modelTemp;
}

This returns a Subject that you can subscribe to.

Answer 3

It depends on what sort of control you need over the resulting observable.

• If you need only to satisfy a requirement that the already-known data be in an observable, then Fateh Mohamed has the right idea in returning a of(this.modelTemp) .

• But if you need to emit future values as the ModelTemp array changes, then German Burgardt is on a better track by providing you a Subject on which you can call behaviorSubject.next(newModelTemp) which will emit to anyone who subscribed to your Behavior Subject . You may also wish to prevent other callers from emitting from this behavior subject by returning only the result of behaviorSubject.asObservable() and keeping the behavior subject itself private.

• Lastly, if the ModelTemp array contains a set of values that should be emitted in succession by the observable, use from(this.modelTemp) .

Answer 4

the method: get(): Observable<ModelTemp> { return of(this.modelTemp); } get(): Observable<ModelTemp> { return of(this.modelTemp); } will reset the error: Type 'Observable<{ name: string; }[]>' is not assignable to type 'Observable<ModelTemp>'. Property 'name' is missing in type '{ name: string; }[]' but required in type 'ModelTemp'. Type 'Observable<{ name: string; }[]>' is not assignable to type 'Observable<ModelTemp>'. Property 'name' is missing in type '{ name: string; }[]' but required in type 'ModelTemp'.