Question

我是rxjs的新手，我需要调用3个异步请求，这些请求最初是作为promises，然后按照调用它们的顺序将重新转换的对象放入数组中。这就是我所拥有的（部分基于this question）：

let observeFront = Observable.fromPromise(this.requestAsync(frontSlice, "front"));
let observeMiddle = Observable.fromPromise(this.requestAsync(midSlice, "middle"));
let observeEnd = Observable.fromPromise(this.requestAsync(endSlice, "end"));

let observeCombined = Observable.concat(observeFront, observeMiddle, observeEnd);

observeCombined.subscribe(data => {
  // this logs only one object - need all 3 in proper order
  console.log(data);
  this.resultsEmitter.emit(data)
});

observeFront.subscribe(response => {
  // logs the correct object
  console.log("front result:", response);
});

observeMiddle.subscribe(response => {
  // correct
  console.log("middle result:", response);
});

observeEnd.subscribe(response => {
  // correct
  console.log("end result:", response);
});

如果我有[frontOb, middleObj, endObj]？

，我怎样才能发出结果？

Answer 1

正如@Eric建议您可以使用forkJoin()代替concat()：

let observeFront = Observable.fromPromise(this.requestAsync(frontSlice, "front"));
let observeMiddle = Observable.fromPromise(this.requestAsync(midSlice, "middle"));
let observeEnd = Observable.fromPromise(this.requestAsync(endSlice, "end"));

let observeJoined = Observable.forkJoin(observeFront, observeMiddle, observeEnd);

observeJoined.subscribe(data => {
  console.log(data); // => [frontOb, middleObj, endObj]
  this.resultsEmitter.emit(data)
});

rxjs：链接异步请求

1 个答案: