Ignore switchMap return value

I want to resolve an observable but I don't want the return value to replace the previous value in the pipe. Is there any asynchronous tap()? I need an operator like a switchMap but I want to ignore the return.

tap()

switchMap

of(1).pipe(switchMap(() => of(2))).subscribe(console.log); // expected: 1


I could create a custom operator but sure there's something built-in in rxjs.

of(1).pipe(switchMap(value => concat(of(2).pipe(ignoreElements()), of(value))))

2 Answers
2

I ended up with this custom operator. It is like tap but resolves observables (and should be updated to also support promises).

export function switchTap<T, R>(next: (x: T) => Observable<R>): MonoTypeOperatorFunction<T>;
export function switchTap<R>(observable: Observable<R>): MonoTypeOperatorFunction<R>;
export function switchTap<T, R>(
arg: Observable<T> | ((x: T) => Observable<R>)
): MonoTypeOperatorFunction<T>
const next: (x: any) => Observable<T


Usage:

of(1).pipe(switchTap(of(2))).subscribe(console.log) // 1


or with a function:

of(1)
.pipe(
switchTap(value =>
console.log(value); // value: 1
return of(value + 1);
)
)
.subscribe(console.log); // 1


If you just want to simply ignore the values of the subscribe, then just don't pass in any arguments in the subscribe callback:

subscribe

of(1).pipe(switchMap(() => of(2))).subscribe(()=>
console.log('no arguments')
);


If you however want to retain the values of the first observable, things can get tricky. One way is to use Subject to retain the value:

Subject

//create a BehaviorSubject
var cache = new BehaviorSubject<any>(0);

of(1).pipe(switchMap((first) =>
cache.next(first);
return of(2);
)).subscribe(() =>
console.log(cache.value) //gives 1
);


Or you can use .map() to alter the values. This is kind of hacky and the code is harder to maintain:

.map()

of(1).pipe(switchMap((first) =>
return of(2).map(() => first);
)).subscribe((second) =>
console.log(second) //gives 1 because the values was mapped
);


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