Question

如何让List<Contributor> Observable<List<Contributor>>中的元素逐个启动，如Observable<Contributor>。 “Contributor”是一个自定义类。

public class MainActivity extends AppCompatActivity {
private TextView mTv;

private interface GitHubService {
    @GET("repos/{owner}/{repo}/contributors")
    Observable<List<Contributor>> contributors(@Path("owner") String owner, @Path("repo") String repo);
}

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    mTv= (TextView) findViewById(R.id.tv_content);

    Retrofit retrofit = new Retrofit.Builder()
            .baseUrl("https://api.github.com")
            .addConverterFactory(GsonConverterFactory.create())
            .addCallAdapterFactory(RxJavaCallAdapterFactory.create())
            .client(new OkHttpClient())
            .build();

    GitHubService service=retrofit.create(GitHubService.class);

    service.contributors("square","retrofit")
            .subscribeOn(Schedulers.io())
            .observeOn(AndroidSchedulers.mainThread())
            .subscribe(new Action1<List<Contributor>>() {
                @Override
                public void call(List<Contributor> contributors) {
                    //TODO
                }
            });
    }
}

这是我的代码。我想最终得到这样的结果：

subscribe(new Action1<Contributor>() {
    @Override
    public void call(Contributor contributor) {
        //TODO 
    }
});

Answer 1

您可以将flatMap应用于Observable<List<...>>并从列表中创建新的observable：

    Observable.just(Arrays.asList("A", "B", "C"))
      .flatMap(new Func1<List<String>, Observable<String>>() {
          @Override
          public Observable<String> call(List<String> list) {
              return Observable.from(list);
          }
      })
      .subscribe(new Action1<String>() {
          @Override
          public void call(String x) {
              System.out.println(x);
          }
      });

如何在Observable <list <contributor>＆gt;的List <contributor>中创建元素？像Observable <贡献者>一样一个一个地推出

1 个答案: