CODEWITHSUNDEEP
typescriptdesign-patternsrxjsobservablevoid
Halt
Halt

Reputation: 2994

Observable<void> usage pattern in Typescript

When an async service has no return value but I want to use Observables I'm inclined to use Observable<boolean>. But I don't have meaning for this boolean value because the service either fails or succeeds and if it fails I want the Observable to be in the error state. This leaves only 'true' values for the observed boolean.

Is the following use of Observable<void> a good pattern for these situations? Or are there foreseeable problems with the use of Observable<void>

const asyncFuncWithoutResult = (): Observable<void> => {
  // fake something async that has no return value

  const itWorked = true; // or not
  if (itWorked) {
    return Observable.of();
  } else {
    return Observable.throw(Error('It did not work'));
  }
}

// call the service
asyncFuncWithoutResult()
  .subscribe(
    undefined, // nothing will be emitted when using Observable.of()
    (err: any) => console.error(err), // handle error state
    () => console.log('Success'),     // handle success state
  );

Upvotes: 22

Views: 15217

Answers (1)

martin
martin

Reputation: 96949

Just to be more precise, when you define a Subject with generic type void there are two ways you can call its next() method and that is without any argument:

const subject = new Subject<void>();
subject.next();

...or with void 0:

subject.next(void 0);

In fact, the value argument to next is optional so if you don't specify it it'll send undefined which means that even if you have Subject<void> you'll still receive next notifications:

asyncFuncWithoutResult().subscribe(alawaysUndefined => {
  /* ... */
});

Also note that you can turn any Observable into <void> (this is necessary when you want to merge multiple Observables) by map:

source.pipe(map(() => void 0))
   ...

Upvotes: 27

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