Question

我目前有这种方法可以正常使用：

  public static Observable<MyCustomObject> run(Service networkService) {
    return Observable.concat(
        networkService.getThingOne().map(response -> {
          Request request = response.raw().request();
          MyCustomObject case = new MyCustomObject(request);
          return case;
        }),
        networkService.getThingOne().map(response -> {
          Request request = response.raw().request();
          MyCustomObject case = new MyCustomObject(request);
          return case;
        }),
        networkService.getThingOne().map(response -> {
          Request request = response.raw().request();
          MyCustomObject case = new MyCustomObject(request);
          return case;
        }),
        networkService.getThingOne().map(response -> {
          Request request = response.raw().request();
          MyCustomObject case = new MyCustomObject(request);
          return case;
        })
    );
  }

一遍又一遍地观察到同样的情况。如果我在那里添加另一个：

  public static Observable<MyCustomObject> run(Service networkService) {
    return Observable.concat(
        networkService.getThingOne().map(response -> {
          Request request = response.raw().request();
          MyCustomObject case = new MyCustomObject(request);
          return case;
        }),
        networkService.getThingOne().map(response -> {
          Request request = response.raw().request();
          MyCustomObject case = new MyCustomObject(request);
          return case;
        }),
        networkService.getThingOne().map(response -> {
          Request request = response.raw().request();
          MyCustomObject case = new MyCustomObject(request);
          return case;
        }),
        networkService.getThingOne().map(response -> {
          Request request = response.raw().request();
          MyCustomObject case = new MyCustomObject(request);
          return case;
        }),
        networkService.getThingOne().map(response -> {
          Request request = response.raw().request();
          MyCustomObject case = new MyCustomObject(request);
          return case;
        })
    );
  }

然后我在concat()下return case下面的红线，但只在case部分。

有人对此有任何想法吗？

编辑：我将问题标题从“为什么不工作”更新为“我怎样才能让这个工作”？我基本上有10个observable，我想坚持使用concat（），我可能会添加更多。所以我需要一些没有限制的东西。

Answer 1

concat()接受并Iterable<T>加上应该合并的观察点：

public static Observable<MyCustomObject> run(Service networkService) {
    ArrayList<Observable<MyCustomObject>> observables = new ArrayList<Observable<MyCustomObject>>();

    observables.add(networkService.getThingOne().map(response -> {
            Request request = response.raw().request();
            MyCustomObject case = new MyCustomObject(request);
            return case;
        }));
    observables.add(networkService.getThingOne().map(response -> {
            Request request = response.raw().request();
            MyCustomObject case = new MyCustomObject(request);
            return case;
        }));
    observables.add(networkService.getThingOne().map(response -> {
            Request request = response.raw().request();
            MyCustomObject case = new MyCustomObject(request);
            return case;
        }));
    observables.add(networkService.getThingOne().map(response -> {
            Request request = response.raw().request();
            MyCustomObject case = new MyCustomObject(request);
            return case;
        });
    observables.add(networkService.getThingOne().map(response -> {
            Request request = response.raw().request();
            MyCustomObject case = new MyCustomObject(request);
            return case;
        }));

    return Observable.concat(observables);
}

请原谅任何语法错误，我暂时没有完成Java

RxJava2如何使用包含4个以上observable的concat（）

1 个答案: