How to delay forkJoin

问题

How would you delay .forkJoin() in rxjs?

Here is what I've got but want to use delay() operator with that?

return forkJoin(
   this.call1(),
   this.call2(),
   this.call3()
 );

So far I got this:

return of(null).pipe(
  delay(5000),
  switchmap(() => this.call1()),
  switchmap(() => this.call2()),
  switchmap(() => this.call3()))
);

That is worked but I would like to use forkJoin, i tried the other soluton

return forkJoin(
   of(this.call1()).pipe(delay(5000)),
   of(this.call2()).pipe(delay(5000)),
   of(this.call3()).pipe(delay(5000))
 );

But it seems like not working.

回答1:

use the delay operator withe the pipe operator

import { delay, take } from 'rxjs/operators';
import { forkJoin } from 'rxjs/observable/forkJoin';
import { of } from 'rxjs/observable/of';

return forkJoin(
   of(call1()).pipe(delay(1000)),
   of(call2()).pipe(delay(2000)),
   of(call3()).pipe(delay(1000))
 );

回答2:

Try thiz

import { delay } from 'rxjs/operators';

return forkjoin(call1(),call2(),call3).pipe(delay(500));

回答3:

Maybe this more complete example will help you find the solution you need:

import { delay } from 'rxjs/operators';
import { Observable } from "rxjs";
import { forkJoin } from 'rxjs/observable/forkJoin';

function call1(): Observable<any> {
  return new Observable<any>(observer => {
    setTimeout(_ => {
      observer.next('CALL1');
      observer.complete();
    }, 1000);
  })
}
function call2(): Observable<any> {
  return new Observable<any>(observer => {
    setTimeout(_ => {
      observer.next('CALL2');
      observer.complete();
    }, 2000);
  });
} 
function call3(): Observable<any> {
  return new Observable<any>(observer => {
    setTimeout(_ => {
      observer.next('CALL3');
      observer.complete();
    }, 3000);
  })
}

forkJoin(call1(), call2(), call3()).pipe(delay(5000)).subscribe(
  response => console.log(response)
)

来源：https://stackoverflow.com/questions/50961190/how-to-delay-forkjoin