如何延迟forkJoin [英] How to delay forkJoin

问题描述

您如何在rxjs中延迟.forkJoin()?

How would you delay .forkJoin() in rxjs?

这是我所需要的，但是想要使用 delay()算什么?

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

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

到目前为止，我知道了:

So far I got this:

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

那是可行的，但是我想使用forkJoin，我尝试了另一种解决方案

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))
 );

但似乎无法正常工作.

推荐答案

将delay运算符与pipe运算符

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))
 );

这篇关于如何延迟forkJoin的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！