RxJS超時不起作用

Question

我試圖在超過3秒時退出Observable流。問題是，當我複製並粘貼相同的值幾次時，distinctUntilChanged運算符不會讓輸入流通過。所以我想超時如果沒有字符串流通過它。這是我的。

 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

超時會拋出TimeoutError，你在處理這個錯誤嗎？

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()結束流：

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