How to call async function inside Observable method sequentially?

Question

I have the following method:

  public classMethod(
    payload: Payload,
  ): Observable<Result> {
    const { targetProp } = payload;
    let target;

    return this.secondClass.secondClassMethod({ targetProp }).pipe(
      delayWhen(() =>
        // some other actions
      ),
    );
  }

It is critical to set this target value before calling this.secondClass.secondClassMethod.

Is it possible to call regular async method like this:

  public classMethod(
    req: classMethodRequest,
  ): Observable<classMethodResponse> {
    const { targetProp } = req;
    let target;

    /**
    **  Calling async method here to set ```target```
    ** like 
    ** target = await someAsyncMethod(targetProp)
    **/

    return this.secondClass.secondClassMethod({ targetProp }).pipe(
      delayWhen(() =>
      ),
      
    );
  }

In other words, I would like to invoke method classMethod so, that classMethod will set target variable within it and then it will be possible to use target in the returning construction

return this.secondClass.secondClassMethod({ targetProp }).pipe(
      delayWhen(() =>
      ),
    );

I've tried to cover async method in:

 from(
      (async () => {
        target = await this.someAsyncMethod.setTarget(targetProp);
      })(),
    );

BUT I was told that this cover will invoke in parallel with

return this.secondClass.secondClassMethod({ targetProp }).pipe(
      delayWhen(() =>
      ),
    );

parallel is not an option here :(

Mrk Sef · Accepted Answer · 2022-07-15 14:37:22Z

3

You can convert the promise to an observable using from and use switchMap

return from(this.someAsyncMethod.setTarget(targetProp)).pipe(
  switchMap(target => this.secondClass.secondClassMethod({ targetProp }),
  delayWhen(() => ... ),
);

edited Jul 15, 2022 at 14:37

Mrk Sef

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answered Jul 15, 2022 at 11:38

Suraj Rao

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2 Comments

watchmecode Over a year ago

Thanks, it works for me. By why just writing async call before return .... will cause that async and code after return will be called in parallel?

Suraj Rao Over a year ago

It shouldn't if you await the call before return

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