Question

您如何在rxjs中延迟.forkJoin()？

这是我所拥有的，但是想要使用delay()运算符吗？

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

到目前为止，我知道了：

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

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

但似乎不起作用。

Answer 1

将delay运算符与pipe运算符一起使用

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))
 );

Answer 2

尝试一下

import { delay } from 'rxjs/operators';

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

Answer 3

也许这个更完整的示例将帮助您找到所需的解决方案：

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)
)

如何延迟forkJoin

3 个答案: