Question

让我们说当s1和o1完成时我需要做，但是s1只发生一次而o1多次发生，每次o1发生时我都需要做一些事情。我怎么能这样做？

我尝试过以下代码，

val s1 = Single.fromCallable {
    Log.d("so","executing s1");
    "hello";
}.cache();

var emitter:ObservableEmitter<String>? = null;
val o1 = Observable.create<String> {
    emitter = it;
}

Observable.zip(s1.toObservable(),o1,
        BiFunction<String, String, Array<String>>() { t1, t2 -> arrayOf(t1,t2); })
        .subscribe { t1 ->
            Log.d("so", "Result = " + t1.joinToString(","));
        }

emitter?.onNext("homer");
emitter?.onNext("marge");
emitter?.onNext("bart");
Log.d("so", "Done");

但结果是

executing s1,
Result = hello,homer
Done

当我预期的结果是

时

executing s1,
Result = hello,homer
Result = hello,marge
Result = hello,bart
Done

Answer 1

使用flatMap：

s1.flatMapObservable({ t1 -> o2.map({ t2 -> arrayOf(t1, t2) }) })

也可以使用PublishSubject而不是偷偷溜出发射器：

val o2 = PublishSubject.create();

s1.flatMapObservable({ t1 -> o2.map({ t2 -> arrayOf(t1, t2) }) })
.subscribe { t1 ->
    Log.d("so", "Result = " + t1.joinToString(","));
}

o2.onNext("homer");
o2.onNext("marge");
o2.onNext("bart");

Log.d("so", "Done");

如何压缩（s1，o1）s1只发生一次而o1多次发生？

1 个答案: