可观察的轮询? [英] Observable polling?

问题描述

我目前有一个可观察的计时器:

I currently have an Observable timer:

private poller(): Observable<PrepareData> {
    return Observable.timer(0, 5000).switchMap(() => this.http.get("/cgi/dashboard.php")).map(res => res.json());
}

我想拥有它，因此在完成最后一个请求后5秒钟完成了一个get请求.有什么简单的方法吗?

I'd like to have it so a get request is done 5 seconds after the last one completed. Any easy way to do this?

推荐答案

这是一个有趣的问题，我很高兴提出解决方案.

This is an interesting question and I had some fun to come up with a solution.

此解决方案如您期望的那样等待上一个完成:

const { Observable, Subject } = Rx

const getFakeHttpRequest$ = () => Observable.of('response !').delay(3000)

const polling$ = new Subject()

Observable
  // need to tick the first time
  .of(null)
  // everytime our polling$ subject will emit, we'll do again what's next
  .merge(polling$)
  // register to a fake request
  .switchMap(_ =>
    getFakeHttpRequest$()
      .do(_ => {
        // once we're here, the request is done
        // no mater how long it was to get a response ...
        console.log('HTTP request done !');

        // ... we wait 5s and then send a new value to polling$ subject
        // in order to trigger a new request
        setTimeout(_ => polling$.next(null), 5000)
      })
  )
  .subscribe()

请注意，我已将请求延迟3秒，并且每8秒就会启动一个新请求，因为我们等待3秒才能得到答复，然后按照您的问题的要求等待5秒.

Notice that I've put a 3s delay on the request, and that a new request kicks in every 8s, as we waited 3s for the response and then 5s as requested in your question.

这是一个正在工作的Plunkr: https://plnkr.co/edit/kXefUbPleqyyUOlfwGuO?p=info

Here's a working Plunkr : https://plnkr.co/edit/kXefUbPleqyyUOlfwGuO?p=info

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