我与我不能改变发送400响应当请求未在API端验证的API进行通信。它是一个有效的HTTP请求，但请求的数据不通过应用程序的验证规则。

400响应包含对为什么请求没有通过验证信息的JSON有效载荷。

我似乎不能，因为Htt的prequestException被抛出，以获得响应主体。有谁知道如何找回这种反应​​的身体吗？

  {尝试        HttpUriRequest请求=参数[0];        HTT presponse serverResponse = mClient.execute（请求）;        BasicResponseHandler处理程序=新BasicResponseHandler（）;        字符串的响应= handler.handleResponse（serverResponse）;        返回响应;    }赶上（HTT presponseException E）{        //投掷了HTTPError这样的        Log.d（TAG的Htt presponseException：+ e.getMessage（））;        Log.d（TAG状态code：+ e.getStatus code（））;        // TODO API成功的HTTP负载返回400，但用户数据无效        如果（e.getStatus code（）== 400）{                //对API错误响应体内信息            }   } 

解决方案

这是我如何发送和获取HTTP响应为字节[] 。当然，你可以改变它的字符串，如果你想要的。

 字节[]缓冲区=新的字节[1024];                HttpClient的=新DefaultHttpClient（）;                httppost =新HttpPost（http://www.rpc.booom.com）;                postParameters =新的ArrayList＆LT;＆的NameValuePair GT;（）;                postParameters.add（新BasicNameValuePair（debug_data，1））;                postParameters.add（新BasicNameValuePair（client_api_ver，1.0.0.0））;                postParameters.add（新BasicNameValuePair（device_identificator，DEVICEID））;                postParameters.add（新BasicNameValuePair（device_resolution）号决议）;                postParameters.add（新BasicNameValuePair（username_hash，hashUser（用户名，密码）））;                postParameters.add（新BasicNameValuePair（password_hash，hashPass（用户名，密码）））;                httppost.setEntity（新UrlEn codedFormEntity（postParameters））;                HTT presponse响应= httpclient.execute（httppost）;                Log.w（响应，状态行：+ response.getStatusLine（）的toString（））;                缓冲液= EntityUtils.toString（response.getEntity（））的getBytes（）; 

希望它帮助！

I'm communicating with an API that I can not change that sends a 400 response when a request is not validated on the API side. It is a valid HTTP request, but the request data does not pass the application's validation rules.

The 400 response contains a JSON payload that has information on why the request did not pass validation.

I can't seem to get the response body because an HttpRequestException is thrown. Does anybody know how to retrieve this response body?

try {
        HttpUriRequest request = params[0];
        HttpResponse serverResponse = mClient.execute(request);

        BasicResponseHandler handler = new BasicResponseHandler();
        String response = handler.handleResponse(serverResponse);
        return response;
    } catch(HttpResponseException e) {
        // Threw HttpError
        Log.d(TAG, "HttpResponseException : " + e.getMessage());
        Log.d(TAG, "Status Code : " + e.getStatusCode());
        // TODO API returns 400 on successful HTTP payload, but invalid user data
        if(e.getStatusCode() == 400) {
                // Information on API error inside Response body
            }
   }

This is how I send and get Http response as an byte[].Of course you can change it to string if you want.

                byte[] buffer = new byte[1024];
                httpclient = new DefaultHttpClient();
                httppost = new HttpPost("http://www.rpc.booom.com");



                postParameters = new ArrayList<NameValuePair>();
                postParameters.add(new BasicNameValuePair("debug_data","1"));
                postParameters.add(new BasicNameValuePair("client_api_ver", "1.0.0.0"));
                postParameters.add(new BasicNameValuePair("device_identificator", deviceId));
                postParameters.add(new BasicNameValuePair("device_resolution", resolution));
                postParameters.add(new BasicNameValuePair("username_hash", hashUser(username,password)));
                postParameters.add(new BasicNameValuePair("password_hash", hashPass(username,password)));

                httppost.setEntity(new UrlEncodedFormEntity(postParameters));

                HttpResponse response = httpclient.execute(httppost);
                Log.w("Response ","Status line : "+ response.getStatusLine().toString());
                buffer = EntityUtils.toString(response.getEntity()).getBytes();

Hope it helps!