使用条件rxjs6重构嵌套订阅

Question

我知道不推荐使用嵌套订阅，但我找不到任何关于重构它们的好资源。

这是我的代码：

this.checkExists(request.id).subscribe(exists => {
  if (exists) {
    this.getDeleteReason(request.id).subscribe(reason => {
      if (reason) {
        this.checkExists(request.id).subscribe(exists2 => {
          if (exists2) {
            // DO DELETE
          }
        });
      }
    });
  }
});

我试着玩过mergeMap，但总是打不开结。

有人能给我指个方向吗？

Answer 1

filter和mergeMap/switchMap的结合。

 this.checkExists(request.id).pipe(
    filter(exists => exists),
    mergeMap(() => this.getDeleteReason(request.id)),
    filter(reason => reason),
    mergeMap(() => this.checkExists(request.id)),
    filter(exist2 => exist2)
  }).subscribe(() => {
      //do delete
  });

编辑:有几种方法可以将reason传递给subscribe。在这种情况下，我会在exist2旁边添加reason。

this.checkExists(request.id).pipe(
    filter(exists => exists),
    mergeMap(() => this.getDeleteReason(request.id)),
    filter(reason => reason),
    mergeMap(reason => ({
             exist2: this.checkExists(request.id), 
             reason})),
    filter(({exist2}) => exist2)
  }).subscribe(({reason}) => {
      //do delete
  });

Answer 2

如果它们是三个独立任务/观察值，那么可以考虑使用forkJoin

let observable1(param1);
let observable2(param2);
let observable3(param3);

let joinedObservables = forkJoin(observable1, observable2, observable3).subscribe(x => {
  let result1 = x[0];
  let result2 = x[1];
  let result3 = x[2];

  ...
});

如果它们的结果相互依赖，你可以使用switchMap，flatMap，mergeMap，exhaustMap (检查差异)

let resultObservable =  return this.observable1().pipe(mergeMap((param1) => {
  return this.observable2().pipe(map((param1) => {

    ....        

    return <result>;
  }));
}));

resultObservable.subscribe(x => {
   ...
});