使用RxJS进行Angular 2轮询 [英] Angular 2 polling with RxJS

问题描述

我正在尝试轮询RESTful端点以刷新我的实时聊天消息.我知道进行实时聊天的最佳方法是Websockets，我只是想了解RxJS如何与Angular 2配合使用.

I'm trying to poll a RESTful endpoint to refresh my live chat messages. I know the best approach for a live chat would be Websockets, I'm just trying to understand how RxJS works with Angular 2.

我想每秒检查一次新消息.我有以下代码:

I want to check for new messages every second. I have the following code:

return Rx.Observable
   .interval(1000)
   .flatMapLatest(() => this.http.get(`${AppSettings.API_ENDPOINT}/messages`))
   .map(response => response.json())
   .map((messages: Object[]) => {
      return messages.map(message => this.parseData(message));
   });

但是我的Typescript编译器返回此错误:

However my Typescript transpiler is returning this error:

类型'Observable< number>'上的属性'flatMapLatest'不存在

Property 'flatMapLatest' does not exist on type 'Observable<number>'

我正在使用RxJS 5.0.0-beta.0

I'm using RxJS 5.0.0-beta.0

如果我使用 merge 而不是 flatMapLatest ，它根本不会调用API.

If I use merge instead of flatMapLatest it doesn't call the API at all.

推荐答案

您需要使用switchMap()，RxJS 5中没有flatMapLatest().

You need to use switchMap(), there's no flatMapLatest() in RxJS 5.

请参见从RxJS 4迁移到5 .尽管文档对switchMap() ...

See Migrating from RxJS 4 to 5... Although docs aren't very clear about switchMap()...

通过将提供的函数应用于源Observable发出的每个项目(返回一个Observable)，然后再发射这些Observable中最近发射的项目，来返回新的Observable.

Returns a new Observable by applying a function that you supply to each item emitted by the source Observable that returns an Observable, and then emitting the items emitted by the most recently emitted of these Observables.

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