如何使用RxJava返回值?

时间:2022-08-27 20:47:09

Let's consider this situation. We have some class which has one method which returns some value:

让我们考虑这种情况。我们有一个类,它有一个方法它返回一些值:

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)
    }
}

How to return that value which was received in onNext? What would be the correct approach? Thank you.

如何返回onNext接收到的值?正确的方法是什么?谢谢你!

1 个解决方案

#1


25  

You need a better understanding of RxJava first, what the Observable -> push model is. This is the solution for reference:

首先,您需要更好地理解RxJava,这是可观察到的>推模型。这是可供参考的解决方案:

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();
  ....
}

#1


25  

You need a better understanding of RxJava first, what the Observable -> push model is. This is the solution for reference:

首先,您需要更好地理解RxJava,这是可观察到的>推模型。这是可供参考的解决方案:

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();
  ....
}