3

您好我正在开发小型 android 应用程序，我想在其中使用带有参数的 HttpUrlConnection 发布请求作为 json 对象。但它对我不起作用，我是通过以下方式做到的：

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。谢谢你 。

我收到以下系统错误

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)
4

2 回答 2

0

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，则会引发此异常。

401 的实际原因是您没有在此时预期的位置发送 OAuth 验证程序代码。

你可以参考下面的代码

    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();
    }
于 2013-06-07T05:41:57.620 回答
0

我强烈建议使用Unirest，它是一个用于发出 http 请求的简单库，它有一个非常易于使用的 api，并且不会过多地使 mem 过载，这是一个关于如何使用您的示例代码的示例。

Unirest.post("https://example.com")
.queryString("places", "['LNCf206KYa5b', 'oWdC0hnm1jjJ']")
.queryString("action", "Do")
.asJson()
于 2015-06-10T19:56:25.903 回答