改进2.0 + RxJava + Error JSON body
改进2.0 + RxJava + Error JSON body
提问于 2015-11-29 07:32:14
我对RxJava和Retrofit非常陌生，正在尝试用它来编写我的应用程序接口调用。所有API调用在出错时都返回JSON主体，其一般格式为，
{"errors":[{"code":100, "message":"Login/Password not valid", "arguments":null}]}目前我的登录API调用的代码(其他也是类似的)是，
mConnect.login(id, password)
.subscribe(new Subscriber<Token>() {
@Override
public void onCompleted() {
Log.d(TAG, "onCompleted()");
}
@Override
public void onError(Throwable e) {
Log.e(TAG, "onError(): " + e);
if (e instanceof HttpException) {
// dump e.response().errorBody()
}
}
@Override
public void onNext(Token token) {
Log.d(TAG, "onNext(): " + token);
}
});当我在onError()中遇到错误时，我希望自动将错误主体中的JSON解码为POJO并使用它。有没有办法做到这一点，最好是在一个地方的所有其他API调用。任何帮助都是非常感谢的。
回答 2
Stack Overflow用户
发布于 2016-03-25 01:37:25
我建议使用可重用的Transformer和onErrorResumeNext运算符来封装您的逻辑。它看起来像这样：
<T> Observable.Transformer<T, T> parseHttpErrors() {
return new Observable.Transformer<T, T>() {
@Override
public Observable<T> call(Observable<T> observable) {
return observable.onErrorResumeNext(new Func1<Throwable, Observable<? extends T>>() {
@Override
public Observable<? extends T> call(Throwable throwable) {
if (throwable instanceof HttpException) {
HttpErrorPojo errorPojo = // deserialize throwable.response().errorBody();
// Here you have two options, one is report this pojo back as error (onError() will be called),
return Observable.error(errorPojo); // in this case HttpErrorPojo would need to inherit from Throwable
// or report this pojo back as part of onNext()
return Observable.just(errorPojo); //in this case HttpErrorPojo would need to inherit from <T>
}
// if not the kind we're interested in, then just report the same error to onError()
return Observable.error(throwable);
}
});
}
};
}注意代码中的注释，因为您必须决定是报告解析后的响应onError()还是onNext()。
然后，您可以在API调用中的任何位置使用此转换器，如下所示：
mConnect.login(id, password)
.compose(this.<Token>parseHttpErrors()) // <-- HERE
.subscribe(new Subscriber<Token>() {
@Override
public void onCompleted() {
Log.d(TAG, "onCompleted()");
}
@Override
public void onError(Throwable e) {
Log.e(TAG, "onError(): " + e);
if (e instanceof HttpErrorPojo) {
// this will be called if errorPojo was reported via Observable.error()
}
}
@Override
public void onNext(Token token) {
Log.d(TAG, "onNext(): " + token);
if (token instanceof HttpErrorPojo) {
// this will be called if errorPojo was reported via Observable.just()
}
}
});Stack Overflow用户
发布于 2016-05-11 21:35:02
反序列化可能也是一个问题。您可以使用retrofit converter to deserialize it (也可以自己动手)。
我的解决方案在murki的基础上增加了一点：
<T> Observable.Transformer<T, T> parseHttpErrors() {
return new Observable.Transformer<T, T>() {
@Override
public Observable<T> call(Observable<T> observable) {
return observable.onErrorResumeNext(new Func1<Throwable, Observable<? extends T>>() {
@Override
public Observable<? extends T> call(Throwable throwable) {
if ( throwable instanceof HttpException ) {
Retrofit retrofit = new Retrofit.Builder()
.baseUrl(SERVER_URL) // write your url here
.addConverterFactory(GsonConverterFactory.create())
.build();
Converter<ResponseBody, Error> errorConverter =
retrofit.responseBodyConverter(Error.class, new Annotation[0]);
// Convert the error body into our Error type.
try {
Error error = errorConverter.convert(((HttpException) throwable).response().errorBody());
// Here you have two options, one is report this pojo back as error (onError() will be called),
return Observable.error(new Throwable(error.getMessage()));
}
catch (Exception e2) {
return Observable.error(new Throwable());
}
}
// if not the kind we're interested in, then just report the same error to onError()
return Observable.error(throwable);
}
});
}
};
}然后在onError()中，
@Override
public void onError(Throwable e) {
progressBar.setVisibility(View.GONE); // optional
if ( !TextUtils.isEmpty(e.getMessage()) ) {
// show error as you like
return;
}
// show a default error if you wish
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/33977686
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