Question

我试图合并3个不同流中的最新值，而忽略其中一个（或多个）尚未发出任何值的最新值。

   const myStream1$:Observable<>=// steam from somewhere,
         myStream2$:Observable<string>=// steam from somewhere,
         myStream3$:Observable<string>=// steam from somewhere;

   Observable.combineLatest(myStream1$, myStream2$, myStream3$)
      .do(([valFromStream1, valFromStream2, valFromStream3])=>
    {
       // I want to do some side-effect operation even if one(or all) of those 3 streams hasn't emit any value yet
    })
    .takeUntil(terminator$)
    .subscribe();

如何实现？

我可以使用下一个逻辑来实现这一点（但是就像sh * t一样丑陋），我想应该有一个运算符可以自动神奇地做到这一点：

   const myStream1$:Observable<string>=// steam from somewhere,
         myStream2$:Observable<string>=// steam from somewhere,
         myStream3$:Observable<string>=// steam from somewhere;

   const subj1 = new BehaviorSubject<string>(undefined),
         subj2 = new BehaviorSubject<string>(undefined),
         subj3 = new BehaviorSubject<string>(undefined);

   myStream1$
      .do((val)=>{
      subj1.next(val);
    }).subscribe();

   myStream2$
      .do((val)=>{
      subj2.next(val);
    }).subscribe();

   myStream3$
      .do((val)=>{
      subj3.next(val);
    }).subscribe();

   Observable.combineLatest(subj1, subj2, subj3)
      .do(([valFromStream1, valFromStream2, valFromStream3])=>
    {
       // I want to do some side-effect operation even if one of those 3 streams hasn't emit any value yet
    })
    .takeUntil(terminator$)
    .subscribe();

Answer 1

您可以强制CombineLatest通过以空值开始每个流来发出，然后可以基于该值是否为空来有条件地执行逻辑

combineLatest(myStream1$.startWith(null),
myStream2$.startWith(null),
myStream3$.startWith(null))
.do(([v1,v2,v3])=>{
....
}).takeUntil(terminator$)

表现为BehaviorSubject的Observable（具有默认值的Observable）

1 个答案: