Question

在服务类中，方法必须传递当前设备的serialNumber。使用ngrx，可以预订device $以获得实际的设备，但这并不好。这种情况下的最佳做法是什么？

@Injectable()
export class DeviceService {
  public device$: Observable<Device>;
  public device: Device; // feels 'redundant'

  constructor(private http: HttpClient,
              private store: Store<DeviceState>,) {
    this.device$ = this.store.select(DeviceSelectors.getCurrentDevice);
    this.device$.subscribe((device: Device) => {
      this.device = device; // feels bad
    });
  }

  regenerate() : Observable<Object> {
    if (environment.production) {
      const url = `${environment.url}/devices/${this.device.serialNumber}/regenerate`;
      return this.http.get(url) as Observable<Object>;
    } else {
      return of({});
    }
  }
}

Answer 1

您可以使用mergeMap实现所需的内容：

@Injectable()
export class DeviceService {
    public device$: Observable<Device>;

    constructor(private http: HttpClient,
                private store: Store<DeviceState>,) {
        this.device$ = this.store.select(DeviceSelectors.getCurrentDevice);
    }

    regenerate() : Observable<Object> {
        if (environment.production) {
            return this.device$
                .pipe(
                    first(),
                    mergeMap(device => this.http.get(`${environment.url}/devices/${device.serialNumber}/regenerate`))
                );
        } else {
            return of({});
        }
    }
}

Answer 2

@NielsSteenbeek，通常我要做的是从存储中即时获取值（调用regenerate（）时），您可以使用flatMap来存档所需的内容

@Injectable()
export class DeviceService {
    constructor(private http: HttpClient, private store: Store<DeviceState>,) {}

    regenerate() : Observable<Object> {
        if (environment.production) {
            return this.store.select(DeviceSelectors.getCurrentDevice)
                .pipe(
                    first(),
                    flatMap(device => this.http.get(`${environment.url}/devices/${device.serialNumber}/regenerate`))
                );
        } else {
            return of({});
        }
    }
}

如何获得可观察的实际价值

2 个答案: