RxJS 连接两个可观察对象

Question

我正在尝试连接两个 Obs1 和 Obs2。 Obs1 发出 5,6,7。 Obs2 发出 1,2,3,4。我正在连接 -> concat(Obs2, Obs1)

我的订阅者期望 1,2,3,4,5,6,7，但只得到 1,2,3,4。我做错了什么？

let Obs1 = new rxjs.Subject();
let Obs2 = new rxjs.Subject();

function sendToObs1(x){ 
    Obs1.next(x)
}

async function sendToObs2(){
  let trns = await getValues();
  for(let i = 0; i < trns.length; i++){
    Obs2.next(trns[i])
  }
  Obs2.complete()
}

function getValues(){
  return new Promise((resolve, reject) => {
    setTimeout(() => resolve([1,2,3,4]), 10)
  })
};

rxjs.concat(Obs2, Obs1).subscribe({
  next: x=> console.log("Received: " + x),
  complete: () => console.log("Done")}
)
sendToObs2()

sendToObs1(5)
sendToObs1(6)
sendToObs1(7)

//Output
// Received: 1 
// Received: 2 
// Received: 3 
// Received: 4


//Expected
// Received: 1 
// Received: 2 
// Received: 3 
// Received: 4
// Received: 5
// Received: 6
// Received: 7

<script src="https://cdnjs.cloudflare.com/ajax/libs/rxjs/6.1.0/rxjs.umd.js"></script>

Answer 1

Concat只在Obs2的onComplete之后订阅

let Obs1 = new rxjs.ReplaySubject();
let Obs2 = new rxjs.Subject();

function sendToObs1(x){ 
    Obs1.next(x)
}

async function sendToObs2(){
  let trns = await getValues();
  for(let i = 0; i < trns.length; i++){
    Obs2.next(trns[i])
  }
  Obs2.complete()
}

function getValues(){
  return new Promise((resolve, reject) => {
    setTimeout(() => resolve([1,2,3,4]), 10)
  })
};

rxjs.concat(Obs2, Obs1).subscribe({
  next: x=> console.log("Received: " + x),
  complete: () => console.log("Done")}
)
sendToObs2()

sendToObs1(5)
sendToObs1(6)
sendToObs1(7)

//Output
// Received: 1 
// Received: 2 
// Received: 3 
// Received: 4


//Expected
// Received: 1 
// Received: 2 
// Received: 3 
// Received: 4
// Received: 5
// Received: 6
// Received: 7

<script src="https://cdnjs.cloudflare.com/ajax/libs/rxjs/6.1.0/rxjs.umd.js"></script>

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