Question

让我们考虑一下这种情况。我们有一些类有一个返回一些值的方法：

public class Foo() {
    Observer<File> fileObserver;
    Observable<File> fileObservable;
    Subscription subscription;

    public File getMeThatThing(String id) {
        // implement logic in Observable<File> and return value which was
        // emitted in onNext(File)
    }
}

如何返回onNext中收到的值？什么是正确的方法？谢谢。

Answer 1

首先需要更好地理解RxJava，Observable是什么 - ＆gt;推模型是。这是供参考的解决方案：

public class Foo {
    public static Observable<File> getMeThatThing(final String id) {
        return Observable.defer(() => {
          try {
            return Observable.just(getFile(id));
          } catch (WhateverException e) {
            return Observable.error(e);
          }
        });
    }
}


//somewhere in the app
public void doingThings(){
  ...
  // Synchronous
  Foo.getMeThatThing(5)
   .subscribe(new OnSubscribed<File>(){
     public void onNext(File file){ // your file }
     public void onComplete(){  }
     public void onError(Throwable t){ // error cases }
  });

  // Asynchronous, each observable subscription does the whole operation from scratch
  Foo.getMeThatThing("5")
   .subscribeOn(Schedulers.newThread())
   .subscribe(new OnSubscribed<File>(){
     public void onNext(File file){ // your file }
     public void onComplete(){  }
     public void onError(Throwable t){ // error cases }
  });

  // Synchronous and Blocking, will run the operation on another thread while the current one is stopped waiting.
  // WARNING, DANGER, NEVER DO IN MAIN/UI THREAD OR YOU MAY FREEZE YOUR APP
  File file = 
  Foo.getMeThatThing("5")
   .subscribeOn(Schedulers.newThread())
   .toBlocking().first();
  ....
}

如何使用RxJava返回值？

1 个答案: