CombineLatest 第一个事件未触发 | 第一个事件未触发

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问题描述

我想知道我的应用程序何时脱机并重新联机.我在 rxjs 中注册了以下事件:

I want to know when my application goes offline and comes back online. I have the following events registered in rxjs:

const online = Rx.Observable.fromEvent(window, 'online');
const offline = Rx.Observable.fromEvent(window, 'offline');

const source = Rx.Observable.combineLatest(online, offline).map(() => navigator.onLine);

source.subscribe(result => {
  console.log('I\'m online: ' + (result ? 'jup' : 'nope'));
});

但是，当我第一次离线时，即使我添加常规的 addEventListener 事件，我也不会触发该事件，但我看到它们被很好地触发了:

But, the first time I go offline the event isn't triggered allthough if I add the regular addEventListener events I see they're getting triggered fine:

window.addEventListener('online', () => console.log('online triggered'));
window.addEventListener('offline', () => console.log('offline triggered'));

以这个jsbin为例，通过devtools和您将在第一次看到它不会登录 我在线:....

Take a look at this jsbin as example, switch of your netwerk via devtools and you'll see the first time it won't log I'm online: ....

推荐答案

combineLatest 操作符要求所有源 Observable 至少发出一个值.

The combineLatest operator requires all source Observables to emit at least one value.

对你来说，这意味着你应该初始化每个流，然后只听变化:

For you this means you should initialize each stream and then just listen to changes:

const source = Rx.Observable.combineLatest(
    online.startWith(null),
    offline.startWith(null),
  )
  .skip(1)
  .map(() => navigator.onLine)

如果你想知道navigator.onLine的初始状态，也许你甚至不需要使用skip(1).

Maybe you don't even need to use the skip(1) if you want to know the initial state of navigator.onLine.

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