RxJs：您可以将运算符作为参数传播到管道运算符中吗

Question

我有两个可观察流，它们执行非常独立的映射逻辑，但最终以以下 3 个运算符结束：

  this.selection
    .pipe(
      ..Custom mapping operators
      tap(_ => this.devicesLoading = true),
      switchMap(d => this.mapService.findLocationForDevices(d)),
      map(loc => marker([loc.latitude, loc.longitude])
    )
    .subscribe(markers => this.plotMarkers(markers));

我想将最后一个tap, switchMap, map运算符移动到一个通用函数，这样我就可以在我的两个可观察流中应用它们。

我想过这样做：

  private resolveLocationsAndConvertToMarkers = (devices: String[]) => [
    tap(_ => this.devicesLoading = true),
    switchMap((devices: string[]) => this.mapService.findLocationForDevices(devices)),
    map(loc => marker([loc.latitude, loc.longitude])
  ];

但我不确定如何将这些运算符扩展到管道参数中，例如：#

      this.selection
        .pipe(
          // Custom mapping operators
          ... this.resolveLocationsAndConvertToMarkers
        )
        .subscribe(markers => this.plotMarkers(markers));

这个错误there are no overloads that expect 3 or 5 arguments...

Answer 1

您可以尝试使用本机 .apply()

this.selection
    .pipe.apply(null,this.resolveLocationsAndConvertToMarkers)

或将运算符列表包装在 pipe()

  private resolveLocationsAndConvertToMarkers = (devices: String[]) => pipe(
    tap(_ => this.devicesLoading = true),
    switchMap((devices: string[]) => this.mapService.findLocationForDevices(devices)),
    map(loc => marker([loc.latitude, loc.longitude])
  );

或返回高阶函数

private resolveLocationsAndConvertToMarkers = (devices: String[]) => source=>source.pipe(
        tap(_ => this.devicesLoading = true),
        switchMap((devices: string[]) => this.mapService.findLocationForDevices(devices)),
        map(loc => marker([loc.latitude, loc.longitude])
      );