Rxjs make sequential calls using concatMap

Question

I would like to make two sequentially calls (if the first one is completed, call the second one):

My Code is like:

        myApiClient
        .firstApiCall()
        .pipe(take(1),
            concatMap (firstResult => {
            doSomethingWithResponse(firstResult);
            return myApiClient.secondApiCall();
        }), take(1))
        .subscribe(secondResult => {
            doSomethingWithResponse(firstResult);
        }, error => {
            catchSecondApiCallError(error);
        });

First question: is that the right way to make sequential calls? Second question: How could I catch the error for the first Call?

What should happen if the first one fails - would you like to continue with the second in that case? — Nicholas K
– Nicholas K, Commented Feb 27, 2021 at 10:36
@NicholasK no, if the first one fails, the second one shouldn't be called (the behaviour now is correct). I want to just handle the error of the first one. — Jimy Weiss
– Jimy Weiss, Commented Feb 27, 2021 at 10:41

Nicholas K · Accepted Answer · 2021-02-27 10:48:06Z

1

Yes, that is the right way to chain sequential api calls using rxjs. For the second question you may make use of the catchError operator.

    myApiClient
    .firstApiCall()
    .pipe(
        take(1),
        catchError(e => {
          console.log('Error caught from first api ', e);
          return EMPTY;
        }),
        concatMap (firstResult => {
          doSomethingWithResponse(firstResult);
          return myApiClient.secondApiCall();
    }), take(1))
    .subscribe(secondResult => {
        doSomethingWithResponse(secondResult);
    }, error => {
        catchSecondApiCallError(error);
    });

EMPTY is imported from:

import { EMPTY } from 'rxjs';

edited Feb 27, 2021 at 10:48

answered Feb 27, 2021 at 10:43

Nicholas K

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2 Comments

Jimy Weiss Over a year ago

thank you. Only for the sake of curiosit: Would be there also a way in order to continue with the second call, in case that the first one would fail?

Nicholas K Over a year ago

You can try using onErrorResumeNext for that.

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