Question

您好我正在开发小型Android应用程序，其中我想使用带有params的HttpUrlConnection post请求作为json对象。但它不适合我，我是按照以下方式做到的：

try 
{
    URL url;
    DataOutputStream printout;
    DataInputStream  input;
    url = new URL ("https://abc.com");
    HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection();
    urlConnection.setRequestMethod("POST");  
    urlConnection.setDoInput (true);
    urlConnection.setDoOutput (true);
    urlConnection.setUseCaches (false);

    urlConnection.setConnectTimeout(10000);  
    urlConnection.setReadTimeout(10000);

    urlConnection.setRequestProperty("Content-Type","application/json");   
    urlConnection.connect();  

    JSONObject jsonParam = new JSONObject();

      JSONArray arr = new JSONArray();
      arr.put("LNCf206KYa5b");
      arr.put("oWdC0hnm1jjJ");
      jsonParam.put("places", arr);
      jsonParam.put("action", "Do");

            printout = new DataOutputStream(urlConnection.getOutputStream ());
            printout.writeUTF(URLEncoder.encode(jsonParam.toString(),"UTF-8"));
            printout.flush ();
            printout.close ();

            int HttpResult =urlConnection.getResponseCode();  

        if(HttpResult ==HttpURLConnection.HTTP_OK){  
        BufferedReader br = new BufferedReader(new InputStreamReader(  
            urlConnection.getInputStream(),"utf-8"));  
        String line = null;  

           while ((line = br.readLine()) != null) {  
            sb.append(line + "\n");  
        }  
        br.close();  

           //System.out.println(""+sb.toString());  

        }else{  
             System.out.println(urlConnection.getResponseMessage());  
        }  
    } catch (MalformedURLException e) {  

        e.printStackTrace();  
    }  
    catch (IOException e) {  

        e.printStackTrace();  
    } catch (JSONException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }finally{  
        if(urlConnection!=null)  
           urlConnection.disconnect();  
    }  
}

它没有给出任何响应代码或任何输出。难道我做错了什么。如何解决这个问题。需要帮忙。谢谢。

我遇到了以下系统错误

06-07 09:55:58.171: W/System.err(4624): java.io.IOException: Received authentication challenge is null
06-07 09:55:58.171: W/System.err(4624):     at org.apache.harmony.luni.internal.net.www.protocol.http.HttpURLConnectionImpl.processAuthHeader(HttpURLConnectionImpl.java:1153)
06-07 09:55:58.171: W/System.err(4624):     at org.apache.harmony.luni.internal.net.www.protocol.http.HttpURLConnectionImpl.processResponseHeaders(HttpURLConnectionImpl.java:1095)
06-07 09:55:58.171: W/System.err(4624):     at org.apache.harmony.luni.internal.net.www.protocol.http.HttpURLConnectionImpl.retrieveResponse(HttpURLConnectionImpl.java:1048)
06-07 09:55:58.171: W/System.err(4624):     at org.apache.harmony.luni.internal.net.www.protocol.http.HttpURLConnectionImpl.getResponseCode(HttpURLConnectionImpl.java:726)
06-07 09:55:58.179: W/System.err(4624):     at org.apache.harmony.luni.internal.net.www.protocol.https.HttpsURLConnectionImpl.getResponseCode(HttpsURLConnectionImpl.java:121)
06-07 09:55:58.179: W/System.err(4624):     at com.mobiotics.qcampaigns.data.operation.ProximityOperation.execute(ProximityOperation.java:187)
06-07 09:55:58.179: W/System.err(4624):     at com.foxykeep.datadroid.service.RequestService.onHandleIntent(RequestService.java:145)
06-07 09:55:58.179: W/System.err(4624):     at com.foxykeep.datadroid.service.MultiThreadedIntentService$IntentRunnable.run(MultiThreadedIntentService.java:170)

Answer 1

由于401响应代码而发生此错误。

您可以像这样检查第一个响应代码

int responsecode = response.getStatusLine().getStatusCode();

如果服务器回复401，则抛出此异常。

您未发送OAuth验证程序代码的401的实际原因。此处预计会出现这种情况。

您可以参考以下代码

    String url = LoginUrl;
    String resultstring = "";

    HttpClient httpclient = new DefaultHttpClient();
    HttpPost httppost = new HttpPost(url);

    // Add your data
    List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(5);
    nameValuePairs.add(new BasicNameValuePair("j_username", username));
    nameValuePairs.add(new BasicNameValuePair("j_password", password));

    try {
        httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

        HttpContext localContext = new BasicHttpContext();
        localContext.setAttribute(ClientContext.COOKIE_STORE,
                Util.cookieStore);

        try {
            HttpResponse response = httpclient.execute(httppost,
                    localContext);

            int responsecode = response.getStatusLine().getStatusCode();
            if (responsecode == 200) {

                Util.responsecode = responsecode;
                resultstring = "Success";
                InputStream in = response.getEntity().getContent();
                resultstring = Util.convertinputStreamToString(in);

            } else if (responsecode == 401) {
                Util.responsecode = responsecode;
            }

        } catch (ClientProtocolException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    }

Answer 2

我强烈建议使用Unirest一个简单的库来制作http请求它有一个非常容易使用的api并且它不会过多地重载内存，这里有一个如何使用它的例子你的示例代码。

Unirest.post("https://example.com")
.queryString("places", "['LNCf206KYa5b', 'oWdC0hnm1jjJ']")
.queryString("action", "Do")
.asJson()

HttpUrlConnection与post请求和参数作为Json对象android

2 个答案: