Question

我试图返回过滤器功能，但返回似乎不适用于回调。这里this.store.let(getIsPersonalized$)是一个可观察的发射布尔值，this.store.let(getPlayerSearchResults$)是一个可观察的发射视频类对象。如何同步运行，我可以完全避免异步回调，因为我无法修改从商店收到的observable。

isPersonalized$ = this.store.let(getIsPersonalized$);
videos$ = this.store.let(getPlayerSearchResults$)                       
                    .map((vids) => this.myFilter(vids));

myFilter(vids) {
   this.isPersonalized$.subscribe((x){
      if(x){
         return this.fileterX(vids);//Return from here
      }
      else {
         return this.filterY(vids);//Or Return from here
      }
  });
}

fileterX(vids) {
  return vids.filter((vid) => vids.views>100;);
}

fileterY(vids) {
  return vids.filter((vid) => vids.views<20;);
}

Answer 1

我以这种方式工作，如果你能在isPersonalized$的订阅上获得分支，你根本不需要myFilter（vids）。这是更新的代码。

  this.store.let(getIsPersonalized$);
  videos$: Observable<any>;

  ngOnInit() {
    this.isPersonalized$.subscribe((x) => {
      if (x) {
        this.videos$ = this.store.let(getPlayerSearchResults$)
                        .map((vids) => this. fileterX(vids));
      } else {
        this.videos$ = this.store.let(getPlayerSearchResults$)
                        .map((vids) => this. fileterY(vids));
      }
    });
  }

  fileterX(vids) {
    return vids.filter((vid) => vids.views>100;);
  }

  fileterY(vids) {
    return vids.filter((vid) => vids.views<20;);
  }

Answer 2

看起来你想在地图函数中评估isPersonalized$的最新值，我会通过withLatestFrom进行评估（示例：第一个每5秒切换一次true/false，第二个每1秒发出越来越多的数字：

const isPersonalized$ = Rx.Observable.interval(5000)
  .map(value => value % 2 === 0);
const getPlayerSearchResults$ = Rx.Observable.interval(1000)
  .withLatestFrom(isPersonalized$)
  .map(bothValues => {
    const searchResult = bothValues[0];
    const isPersonalized = bothValues[1];
    ...
  });

如何从观察者中返回？

2 个答案: