首页 > Web开发 > 详细

[RxJS] Flatten a higher order observable with concatAll in RxJS

时间:2016-12-16 22:49:02      阅读:207      评论:0      收藏:0      [点我收藏+]

Besides switch and mergeAll, RxJS also provides concatAll as a flattening operator. In this lesson we will see how concatAll handles concurrent inner Observables and how it is just mergeAll(1).

 

const clickObservable = Rx.Observable
  .fromEvent(document, click);

const clockObservable = clickObservable
  .map(click => Rx.Observable.interval(1000).take(5))
  .concatAll(); // the same as .mergeAll(1)

// flattening
// Observable<Observable<number>> ---> Observable<number>

/*
--------+--------------+-+----
        \        
         -0-1-2-3-4|
         
         concatAll
         
----------0-1-2-3-4-----0-1-2-3-4--0-1-2-3-4
*/

clockObservable
  .subscribe(x => console.log(x));

 

[RxJS] Flatten a higher order observable with concatAll in RxJS

原文:http://www.cnblogs.com/Answer1215/p/6188384.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!