RxJS combineLatest混淆

Question

我以为我理解combineLatest，但考虑到我的输出-我不理解它。我认为，在combineLatest中，所有可观测值都会在任何可观测值发出时发出它们的最后值。

(注:我只是采取了(5)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}
  `)
})

我以为能看到

"
      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

但我得到了

"
      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。你的想法，为什么我没有看到什么，我的期望将是可怕的！

Answer 1

http://reactivex.io/documentation/operators/combinelatest.html CombineLatest在任何源观测者发出一个项目时都会发出一个项目。

(只要每个源可观测数据至少发出一个项) <<==

int3不会在3000 is之前发出任何项，因此Rx将等待它，然后用最新的项调用onNext

可能的解决方案:尝试使用计时器(在0发出第一个值，然后每n次发出一次)。

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}
  `)
})