filter out observableB if observalbeA get triggered 1s before

Question

Let's say I have two observables, observableA and observableB.

There are 3 cases for how these two observables can be triggered:

1. Only observableA gets triggered.
2. Only observableB gets triggered.
3. ObservableA gets triggered first, followed by observableB after a 1-second delay.

How can I handle these two observables so that the callback function won't execute twice?

expecting: the callback function won't execute twice

Answer 1

So what's not said is that it's ok to wait to see if B is going to trigger after an A. The only way to solve this is have a window in which after A triggers you wait X seconds then trigger to continue with A, or you receive a B and go with that logic. bufferTime can serve as this. First we merge A and B together then we buffer those 2 observables up.

import { interval, bufferTime, merge, map, switchMap, timer } from 'rxjs';

const obsA = interval(1000)
             .pipe( switchMap( (i) => timer( Math.random() * 1000 ) ) )
             .pipe( map( (x) => `A`) )
const obsB = interval(1000)
             .pipe( switchMap( (i) => timer( Math.random() * 1000 ) ) )
             .pipe( map((y) => `B`))

merge( 
  obsA, 
  obsB 
).pipe( bufferTime(1000) )
.subscribe(x => {
  if( x.length === 0 ) return;
  if( x.every( (item) => item.startsWith("A") ) ) {
    console.log("Only A", x);
  } else if( x.findIndex( (item) => item.startsWith("B") ) === 0 ) {
    console.log("Only B", x);
  } else {
    console.log("A followed by B", x);
  }
})

You have to handle the case where neither A nor B fire which will be an empty array, and just ignore it.

Answer 2

Here is the example for you, once the first observable is triggered, delay 1s and switchmap triggers the second once.

import { of } from 'rxjs';
import { delay, switchMap } from 'rxjs/operators';

const firstObservable = of('Hello');
const secondObservable = of('World');

firstObservable.pipe(
  delay(1000),
  switchMap(() => secondObservable)
).subscribe(result => console.log(result));