RxJS连接了两个可观测值

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。我正在凹陷->concat(Obs2，Obs1(

在我的订阅者中期望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仅在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>