CODEWITHSUNDEEP
rxjs
user2765977
user2765977

Reputation: 499

RXjs emit a value to a subscriber before piping another value to the same subscriber

I am trying to do the following:

observable.pipe(
  emit-value-to-subscriber(false),
  map((value) => doSomethingWithValue(value)),
).subscribe(console.log);

Essentially intercept the value from observable and first emit a false value to the subscriber before piping the processed version of the value.

Is there a good way to do this in RXjs? My current implementation just involves using two subscribers:

observable.pipe(
  map(() => false),
).subscribe(console.log);

observable.pipe(
  map((value) => doSomethingWithValue(value)),
).subscribe(console.log);

But I have serious doubts about the reliability of doing it this way (order of events).

Upvotes: 0

Views: 2355

Answers (2)

stck
stck

Reputation: 82

Solution without concat and startsWith - if your want to handle a stream, e.g. false, true, false, false, ..., you will get a false, %mod false%, true, %mod true% ....

Observable.from(false)
  .pipe(
    flatMap(x$ => from([x$, doSomethingWithAValue(x$)])),
  )

Upvotes: 1

Andrei Tătar
Andrei Tătar

Reputation: 8295

There are multiple ways of doing what you want.

Assuming doSomethingWithValue returns a boolean, you can use startWith operator:

observable.pipe(
  map((value) => doSomethingWithValue(value)),
  startWith(false),
).subscribe(console.log);

you can also use concat:

concat(
  of(false),
  observable.pipe(map(value => doSomethingWithValue(value)),
).subscribe(console.log);

you can also use concat inside switchMap:

observable.pipe(
  switchMap(value => concat(of(false), of(doSomethingWithValue(value)))),
).subscribe(console.log);

if there's a different observer, you can use tap:

observable.pipe(
  tap(_ => observer.next(false)),
  map((value) => doSomethingWithValue(value)),
).subscribe(console.log);

Upvotes: 3

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