Javascript RxJS CombinelateTest运算符异常行为

我有一个关于RxJS CombineTest运算符的查询。我已修改了中给出的示例

详情如下:

//timerOne emits first value at 1s, then once every 4s
const timerOne = Rx.Observable.timer(1000, 4000);
//timerTwo emits first value at 2s, then once every 4s
const timerTwo = Rx.Observable.timer(2000, 4000)
//timerThree emits first value at 3s, then once every 4s
const timerThree = Rx.Observable.of(false);

//when one timer emits, emit the latest values from each timer as an array
const combined = Rx.Observable
.combineLatest(
    timerOne,
    timerTwo,
    timerThree
);

const subscribe = combined.subscribe(latestValues => {
    //grab latest emitted values for timers one, two, and three
    const [timerValOne, timerValTwo, timerValThree] = latestValues;


  if(latestValues[0] === 3) {    
    this.timerThree = Rx.Observable.of(true);
  }

  console.log(
    `Timer One Latest: ${timerValOne}, 
     Timer Two Latest: ${timerValTwo}, 
     Timer Three Latest: ${timerValThree}`
   );
});

我希望TimerTree的值更改为true位，它始终保持打印false，如输出片段所示：

"Timer One Latest: 3, 
 Timer Two Latest: 2, 
 Timer Three Latest: false"
"Timer One Latest: 3, 
 Timer Two Latest: 3, 
 Timer Three Latest: false"
"Timer One Latest: 4, 
 Timer Two Latest: 3, 
 Timer Three Latest: false"

---

知道为什么会这样吗？有办法解决这个问题吗？感谢

这里需要注意的重要一点是，

TimerTree

本身不是一个可观察的对象，而是对可观察对象的引用。使用

combinelatetest

时，它是在组合该对象，而不是引用该对象的变量。因此，当您将

timerTree

指定给一个新的可观察对象时，它现在指向一个新对象，但

组合的

仍在使用旧对象

---

如果您希望能够更改

TimerTree

的值，请尝试使用

主题

。然后，您可以使用

TimerTree向其推送新值。接下来

补充John的答案：

this.timerTree

在

最新值[0]==3

时未定义，因为在lambda函数内部时，

this

指的是“最近的外部范围”

如果要在浏览器中运行此操作，则

此

将是

窗口

对象，因此您只需向窗口对象添加属性

另外，

timerTree

被定义为

const

，这意味着如果您尝试对同一对象进行重新分配（但如上所述，您正在分配给不同的对象），它将抛出错误

通过摆弄fiddle，我得到了一些您想要的东西，尽管这需要工作来消除代码重复：

//timerOne emits first value at 1s, then once every 4s
const timerOne = Rx.Observable.timer(1000, 4000);
//timerTwo emits first value at 2s, then once every 4s
const timerTwo = Rx.Observable.timer(2000, 4000)
//timerThree emits first value at 3s, then once every 4s
let timerThree = Rx.Observable.timer(3000, 4000)

//when one timer emits, emit the latest values from each timer as an array
let combined = Rx.Observable
.combineLatest(
    timerOne,
    timerTwo,
    timerThree
);

const subscribe = combined.subscribe(latestValues => {
    //grab latest emitted values for timers one, two, and three
    const [timerValOne, timerValTwo, timerValThree] = latestValues;

  if(latestValues[0] === 3) {
    console.log("this ===>", this);
    console.log("this.timerThree ===> ", this.timerThree);
    subscribe.unsubscribe();
    combined = Rx.Observable.combineLatest(timerOne, timerTwo, Rx.Observable.of(true));
    combined.subscribe(lvs => {
        const [tv1, tv2, tv3] = lvs
      console.log(
        `Timer One Latest: ${tv1}, 
         Timer Two Latest: ${tv2}, 
         Timer Three Latest: ${tv3}`
       );
    })
  }
  console.log(
    `Timer One Latest: ${timerValOne}, 
     Timer Two Latest: ${timerValTwo}, 
     Timer Three Latest: ${timerValThree}`
   );
});

请注意

unsubscribe（）

调用以防止以前组合的计时器再次执行，新调用

combinelateest

和新的

observeable

以打印

true

我还必须将

timerTree

从

const

更改为

let

，以便能够重新分配它