如何在 RxJava 出错后更改重试请求中的参数

Question

我向带有改造 2.0 的服务器发送登录请求，并向客户端会话令牌发送服务器 return，我必须在其他请求中使用它，但此令牌的生命周期有限，并且何时可用过期服务器 returns HTTP 错误 401。

我尝试重新登录，在收到此错误后，在下一个代码的帮助下：

    holder.getApi(GuideProfileApi.class)
      .getProfile(String.valueOf(holder.getServerId()), holder.getServerToken())
      .subscribeOn(Schedulers.io())
      .retryWhen(new Function<Observable<Throwable>, ObservableSource<?>>() {
         @Override
         public ObservableSource<?> apply(Observable<Throwable> throwableObservable) throws Exception {
           return throwableObservable.flatMap(new Function<Throwable, ObservableSource<?>>() {
             @Override
             public ObservableSource<?> apply(Throwable throwable) throws Exception {
               if (throwable instanceof HttpException &&  ((HttpException)throwable).code() == 401) {
                 RegistryLoginResult loginResult = holder.login().blockingSingle();
                 return holder.getApi(GuideProfileApi.class)
                    .getProfile(String.valueOf(loginResult.getUserId()), loginResult.getSessionToken());
               }
               return Observable.error(throwable);
            }
        });
     }
  })
  .observeOn(AndroidSchedulers.mainThread())
  .subscribe(new Consumer<ProfileResult>() {
    @Override
    public void accept(ProfileResult profileResult) throws Exception {
      Log.d("Result", profileResult.toString());
    }
  }, new Consumer<Throwable>() {
      @Override
      public void accept(Throwable throwable) throws Exception {
        Log.e("Result", throwable.getLocalizedMessage());
      }
});

并发送重试请求，但请求参数与错误请求（重新登录前）相同。如何在再次发送之前更改请求的参数？

Answer 1

您用错了运算符。如果遇到错误，retryWhen 将重试您的 原始可观察对象 。你需要的是onErrorResumeNext。像

 holder.getApi(GuideProfileApi.class)
      .getProfile(String.valueOf(holder.getServerId()), holder.getServerToken())
      .subscribeOn(Schedulers.io())
      .onErrorResumeNext(new Function<Throwable, ObservableSource<?>>() {
        @Override
         public ObservableSource<?> apply(Throwable throwable) {
               if (throwable instanceof HttpException &&  ((HttpException)throwable).code() == 401) {
                 RegistryLoginResult loginResult = holder.login().blockingSingle();
                 return holder.getApi(GuideProfileApi.class)
                    .getProfile(String.valueOf(loginResult.getUserId()), loginResult.getSessionToken());
               }
               return Observable.error(throwable);
        }
    }) 
  .observeOn(AndroidSchedulers.mainThread())
  .subscribe(new Consumer<ProfileResult>() {
    @Override
    public void accept(ProfileResult profileResult) throws Exception {
      Log.d("Result", profileResult.toString());
    }
  }, new Consumer<Throwable>() {
      @Override
      public void accept(Throwable throwable) throws Exception {
        Log.e("Result", throwable.getLocalizedMessage());
      }
});

Answer 2

您可以使用retryWhen，但问题是您的retryWhen 会重试您在惰性时刻创建的同一个可观察对象。您在这里的解决方案是使用运算符 defer 来获取 host()，因为 defer 它不是在您定义它时创建可观察对象，而是在它被订阅者使用时创建。

Observable.defer(()-> holder.getApi(GuideProfileApi.class)
          .getProfile(String.valueOf(holder.getServerId()),holder.getServerToken()))
  .subscribeOn(Schedulers.io())
  .retryWhen(new Function<Observable<Throwable>, ObservableSource<?>>() {
     @Override
     public ObservableSource<?> apply(Observable<Throwable> throwableObservable) throws Exception {
       return throwableObservable.flatMap(new Function<Throwable, ObservableSource<?>>() {
         @Override
         public ObservableSource<?> apply(Throwable throwable) throws Exception {
           if (throwable instanceof HttpException &&  ((HttpException)throwable).code() == 401) {
             RegistryLoginResult loginResult = holder.login().blockingSingle();
             return holder.getApi(GuideProfileApi.class)
                .getProfile(String.valueOf(loginResult.getUserId()), loginResult.getSessionToken());
           }
           return Observable.error(throwable);
        }
    });
 }
  })
  .observeOn(AndroidSchedulers.mainThread())
  .subscribe(new Consumer<ProfileResult>() {
    @Override
    public void accept(ProfileResult profileResult) throws Exception {
      Log.d("Result", profileResult.toString());
    }
      }, new Consumer<Throwable>() {
          @Override
          public void accept(Throwable throwable) throws Exception {
            Log.e("Result", throwable.getLocalizedMessage());
          }
    });

你可以在这里看到一些重试的例子https://github.com/politrons/reactive/blob/master/src/test/java/rx/observables/errors/ObservableExceptions.java

如何在 RxJava 出错后更改重试请求中的参数

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