Question

我以为我理解combineLatest，但考虑到我的输出 - 我不理解它。我想用combineLatest，只要任何一个可观察者发射，所有可观察者都会发出它们的最后一个值。

（注意：我刚刚执行了take（5）来限制我的控制台输出）

所以，考虑到这个微不足道的例子 -

const int1$ = Rx.Observable.interval(1000).take(5)
const int2$ = Rx.Observable.interval(500).take(5)
const int3$ = Rx.Observable.interval(3000).take(5)
const all$ = Rx.Observable.combineLatest(
  int1$, int2$, int3$ 
)

all$.subscribe(latestValues => {
  const [int1, int2, int3] = latestValues;
  console.log(`
      interval one @ 1000 ${int1},
      interval two @ 500 ${int2},
      interval three @ 3000 ${int3}
  `)
})

我想看看

"
      interval one @ 1000 0,
      interval two @ 500 1,
      interval three @ 3000 0
  "
"
      interval one @ 1000 1,
      interval two @ 500 2,
      interval three @ 3000 0
  "
"
      interval one @ 1000 1,
      interval two @ 500 3,
      interval three @ 3000 1
  "
"
      interval one @ 1000 2,
      interval two @ 500 4,
      interval three @ 3000 1

但我正在

"
      interval one @ 1000 2,
      interval two @ 500 4,
      interval three @ 3000 0
  "
"
      interval one @ 1000 3,
      interval two @ 500 4,
      interval three @ 3000 0
  "
"
      interval one @ 1000 4,
      interval two @ 500 4,
      interval three @ 3000 0
  "
"
      interval one @ 1000 4,
      interval two @ 500 4,
      interval three @ 3000 1

有点困惑。你对我为什么没有看到我期望的想法会很棒！

Answer 1

http://reactivex.io/documentation/operators/combinelatest.html 只要任何源Observables发出项目

，CombineLatest就会发出一个项目

（只要每个源Observable发出至少一个项目）＆lt;＆lt; == THIS

int3在3000ms之前没有发出任何项目，所以Rx等待它，然后使用最新项目调用onNext

可能的解决方案：尝试使用定时器（在0秒时发出第一个值，然后每n秒发出一次）

const int1$ = Rx.Observable.timer(0,1000).take(5)
const int2$ = Rx.Observable.timer(0,500).take(5)
const int3$ = Rx.Observable.timer(0,3000).take(5)
const all$ = Rx.Observable.combineLatest(
  int1$, int2$, int3$ 
)

all$.subscribe(latestValues => {
  const [int1, int2, int3] = latestValues;
  console.log(`
      interval one @ 1000 ${int1},
      interval two @ 500 ${int2},
      interval three @ 3000 ${int3}
  `)
})

RxJS结合最大的混乱

1 个答案: