Angular 2-Observable的排序列表

问题描述

sort来自Observable的项目列表并且仍然能够使用async pipe的最佳方法是什么? (我读过，创建自定义排序管道并不是真正有效.)我想避免订阅和保留数据的本地副本，因此仅使用异步管道...

What is the best way to sort a list of items coming from an Observable and still be able to use the async pipe? (I read that making a custom sort pipe is not really efficient.) I want to avoid subscribing and keeping a local copy of data and thus just using async pipe...

//can I use map here and filter items right in the observable and get rid of subscribe?

this.test$ = Observable.of(['one', 'two', 'three'])
    .subscribe((data) => {
        data.sort((a, b) => {
            return a < b ? -1 : 1;
         });
        this.data = data;
     });

模板:

<div *ngFor='let item of data'>
<!-- want to be able to use async pipe here -->

推荐答案

如果调用.subscribe()，您将得到一个Subscription，则异步管道需要一个Observable.

If you call .subscribe() you get a Subscription, the async pipe expects an Observable.

如果将其更改为

this.test$ = Observable.of(['one', 'two', 'three'])
.map((data) => {
    data.sort((a, b) => {
        return a < b ? -1 : 1;
     });
    return data;
 });

您可以将异步管道与test$

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