Question

问题很简单：如何从Observable<Array<T>>转换为Observable<T>？

如果您能用rxJS或RxJava写答案，我会很感激，但其他任何语言都很好。

Answer 1

根据您的预期结果，有多种方法可以解决您的问题。这是一个fiddle：

console.clear();

var x = Rx.Observable.of([1, 2, 3], [4, 5, 6]);

x.subscribe(item => {
  console.log('Without flatMap: ' + JSON.stringify(item));
  //Without flatMap: [1, 2, 3]
  //Without flatMap: [4, 5, 6]
});

x.flatMap(item => {
    return item;
  }
).subscribe(item => {
  console.log('With flatMap: ' + item);
  //With flatMap: 1
  //With flatMap: 2
  //With flatMap: 3
  //With flatMap: 4
  //With flatMap: 5
  //With flatMap: 6
});

x.reduce((a, b) => {
  return a.concat(b);
}).subscribe(item => {
  console.log('With reduce: ' + JSON.stringify(item));
  //With reduce: [1, 2, 3, 4, 5, 6]
});

x.scan((a, b) => {
  return a.concat(b);
}).subscribe(item => {
  console.log('With scan: ' + JSON.stringify(item));
  //With scan: [1, 2, 3]
  //With scan: [1, 2, 3, 4, 5, 6]
});

Answer 2

对于Java，Observable.from()就是你想要的：

Observable<List<String>> listObservable = Observable.just(Arrays.asList("why", "hello", "there"));

listObservable.flatMap(Observable::from)
              .subscribe(str -> System.out.print(str + "-"),
                         throwable -> {});

可观察到的<阵列<T>＆GT;到Observable <t>

2 个答案: