CODEWITHSUNDEEP
rxjs
Larest
Larest

Reputation: 386

How make an observable buffer data till an event happen and stop buffering?

I have outward stream of drawing from an user and I need wait till data of the user profile come.

So in general I want buffer data till an event happens and then play this data and skip buffering.

I can do it with external array like the follow code show (stackblitz):


    import { of, interval } from 'rxjs'; 
    import { map, take } from 'rxjs/operators';

    const numbers = interval(1000);
    const source = numbers.pipe(
      take(4)
    );
    let allowPerform = false;
    setTimeout(_=>{allowPerform = true},2001);
    const fifo=[];
    source.subscribe(x => {
      fifo.push(x);
      if (!allowPerform) {
         console.log('skip perform for:', x);
         return; 
      }

      let item ;  
      while ( fifo.length) {
        item = fifo.shift(); 
        console.log('perform for:', item);
      } 
    });

it's output:


skip perform for: 0
skip perform for: 1
perform for: 0
perform for: 1
perform for: 2
perform for: 3

However how to do it in RXjs's way?

Upvotes: 1

Views: 741

Answers (1)

Andrei Gătej
Andrei Gătej

Reputation: 11934

Here could be a way to do this:

// Buffer until the `notifier` emits
// Then do not buffer anymore, just send the values

const src$ = interval(1000).pipe(take(10), publish(), refCount());
const notifier$ = timer(2001);

concat(
  src$.pipe(buffer(notifier$), mergeAll()),
  src$
).subscribe(console.log, null, () => console.warn('complete'));

Using publish(), refCount() will multicast the emitted values to all the consumers. This is achieved by placing a Subject between the source and the data consumers, meaning that the source won't be subscribed multiple times.

src$.pipe(buffer(notifier$), mergeAll()), will buffer until notifier$ emits. But, because notifier$ also completes, the entire observable passed will complete, allowing for the next observable(src$) to be subscribed.
mergeAll() is used because buffer will emit an array of the collected values and with mergeAll() we can get the values separately.

StackBlitz

Upvotes: 2

Related Questions