展平嵌套订阅

Question

鉴于以下内容：

this.activatedRoute.parent.params
  .subscribe(params => {
    let fooId = +params['fooId'];

    this.activatedRoute.params
      .subscribe(params => {
        let barId = +params['barId'];

        this.fooBarService.getFooBars(fooId, barId)
          .subscribe(fooBars => {
            this.fooBars = fooBars;
          });
      });
  });

可以使用以下内容简化部分嵌套：

this.activatedRoute.parent.params
  .subscribe(params => {
    let foo = +params['fooId'];

    this.activatedRoute.params
      .switchMap(params => this.fooBarService.getFooBars(fooId, +params['barId']))
      .subscribe(fooBars => {
        this.fooBars = fooBars;
      });
  });

但是可以展平剩余的嵌套订阅吗？

Answer 1

使用combineLatest运算符：

Observable
  .combineLatest(this.activatedRoute.parent.params, this.activatedRoute.params)
  .switchMap(data => this.fooBarService.getFooBars(+(data[0])['fooId'], +(data[1])['barId']))
  .subscribe(fooBars => {
    this.fooBars = fooBars;
  });