RxJS超时不起作用

Question

我试图在超过3秒时退出Observable流。问题是，当我复制并粘贴相同的值几次时，distinctUntilChanged运算符不会让输入流通过。所以我想超时如果没有字符串流通过它。这是我的。

 import { Subject } from "rxjs/Subject"; 
     import "rxjs/add/operator/filter"; 
     import "rxjs/add/operator/debounceTime"; 
     import "rxjs/add/operator/distinctUntilChanged"; 
     import "rxjs/add/operator/switchMap"; 
     import "rxjs/add/operator/timeout"; 

     this._searchSubject 
     .filter(val => val.length > 0) 
     .debounceTime(500) 
     .distinctUntilChanged() 
     .timeout(3000) 
     .switchMap(userSearchInput => { 
      ...api call that returns Promise 
     }) 
     .subscribe(searchResults => { 
      ...do stuff with the result 
     }); 

Answer 1

超时会抛出TimeoutError，你在处理这个错误吗？

Rx.Observable.from(new Promise(resolve => setTimeout(resolve, 1000))) 
    .timeout(500) 
    .subscribe(console.log, ({ message }) => console.error(message)); 

或者您可以使用timeoutWith & Rx.Observable.empty()结束流：

Rx.Observable.from(new Promise(resolve => setTimeout(resolve, 1000))) 
    .timeoutWith(500, Rx.Observable.empty()) 
    .subscribe(null, null,() => console.log('done'));