不调用带有ReplaySubject的forkJoin.subscribe

Question

我想创建一个Observable，如果两个ReplaySubject实例都持有一个值，则将触发它。我这样尝试过：

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同时调用let v1 = new Rx.ReplaySubject(); let v2 = new Rx.ReplaySubject(); v1.next('X'); const combo = Rx.Observable.forkJoin(v1, v2) combo.subscribe( arr => console.log(arr) ) v1.subscribe( s => console.log('v1', s) ); v2.subscribe( s => console.log('v2', s) ); v2.next('Y'); 和v1.subscribe，但不调用v2.subscribe。我怎么了如何解决此问题？

Answer 1

forkJoin等待输入可观察值完成，然后发出最后发出的项。您必须完成两个主题。

Answer 2

forkJoin将仅在输入可观察值全部完成后才发出所有最后一个值的数组。如果您手动完成v1和v2，则combo将发出：

let v1 = new rxjs.ReplaySubject();
let v2 = new rxjs.ReplaySubject();

v1.next('X');
v2.next('Y');
v2.next('Z'); // This additional value should demonstrate the idea of forkJoin

const combo = rxjs.forkJoin(v1, v2)
combo.subscribe( arr => console.log(arr) )

v1.complete();
v2.complete();

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如果您不希望内部观察对象完成操作，则可能会寻找combineLatest：

let v1 = new rxjs.ReplaySubject();
let v2 = new rxjs.ReplaySubject();

v1.next('X');
v2.next('Y');

const combo = rxjs.combineLatest(v1, v2)
combo.subscribe( arr => console.log(arr) )

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