Question

我正在使用RxAndroid，并且由于两个observable，我想执行一个改装请求。所以我使用的是zip运算符

Observable<RequestBuilder> obs = a.zipWith(b, (lhs, rhs) -> builder.b(rhs).a(lhs));

现在问题在于builder.build（）返回另一个可观察的 ObservableRequest ，但是这个方法现在使它成为 ObservableObservableRequest ，我似乎无法弄清楚。我已经这样做了，但我想正确地做到这一点......

obs.map(
(builder) -> builder.build(api)
.subscribeOn(Schedulers.newThread())
.observeOn(AndroidSchedulers.mainThread())
.subscribe(this::setData)
);

也许像

    obs.map((builder) -> builder.build(api))
.subscribeOn(Schedulers.newThread())
.observeOn(AndroidSchedulers.mainThread())
.subscribe(this::setData);

Answer 1

DeBuGGeR，这可能会有所帮助。

login类

中的

RestPresenter方法

public void login(String email, String password) {
        mRestService.getApi()
                .postAuth(email, password, "") //create Observable
                .timeout(TIMEOUT_IN_SECONDS, TimeUnit.SECONDS) //time out
                .retry(RETRY_COUNT_FOR_REQUEST) //times to retry
                .subscribeOn(Schedulers.newThread())
                .observeOn(AndroidSchedulers.mainThread())
                .subscribe(new Observer<AuthBean>() { // your model
                    @Override
                    public void onCompleted() {
                        // do nothing
                    }

                    @Override
                    public void onError(Throwable e) {
                        mView.showError("Error");
                    }

                    @Override
                    public void onNext(AuthBean posts) {
                        mView.workWithLoginResponse(posts);
                    }
                });
}

在activity

中调用它

RestService mRestService = new RestService();
mRestPresenter = new RestPresenter(YourActivity.this, mRestService);
mRestPresenter.login(email, password);

RestService类可能看起来像这样

public class RestService {
    private ProviderApi mProviderApi;
    private long CONNECTION_TIMEOUT_IN_SECONDS = 15L;
    private long READ_TIMEOUT_IN_SECONDS       = 15L;
    private long WRITE_TIMEOUT_IN_SECONDS      = 25L;

    public RestService() {
        RequestInterceptor requestInterceptor = new RequestInterceptor() {
            @Override
            public void intercept(RequestFacade request) {
                request.addHeader("Accept", "application/json");
            }
        };
        RestAdapter restAdapter = new RestAdapter.Builder()
                .setEndpoint(Config.URL_SERVER)
                .setRequestInterceptor(requestInterceptor)
                .setLogLevel(RestAdapter.LogLevel.FULL)
                .setClient(new OkClient(getClient())) // need square/okhttp
                .build();
        mProviderApi = restAdapter.create(ProviderApi.class);
    }

    private OkHttpClient getClient() {
        OkHttpClient client = new OkHttpClient();
        client.setConnectTimeout(CONNECTION_TIMEOUT_IN_SECONDS, TimeUnit.MINUTES);
        client.setReadTimeout(READ_TIMEOUT_IN_SECONDS, TimeUnit.MINUTES);
        client.setWriteTimeout(WRITE_TIMEOUT_IN_SECONDS, TimeUnit.MINUTES);
        return client;
    }

    public ProviderApi getApi() {
        return mProviderApi;
    }
}

mView是您的activity，因此您只需创建workWithLoginResponse方法，并通过服务器答案执行您想要的操作。

此外，ProviderApi是界面，可能包含以下内容：

@FormUrlEncoded
@POST("/auth")
Observable<AuthBean>
postAuth(@Field("login") String login,
         @Field("password") String password);

Answer 2

正在返回ObservableObservableRequest，但我想要 ObservableRequest将被退回

flatMap怎么样？

//create Observable<Observable<Integer>>
Observable.just(Observable.just(1, 2, 3))
        //convert to Observable<Integer>
        .flatMap(integerObservable -> integerObservable)
        //receive Integer values
        .subscribe(integer -> {
            //do something with Integer
        });

More about flatMap

使用RxJava进行改造时处理额外的可观察性

2 个答案: