HttpUrlConnection，包含发布请求和参数，作为Json对象android

8ehkhllq 于 2023-01-03 发布在 Android

关注(0)|答案(2)|浏览(140)

嗨，我正在开发一个小的Android应用程序，我想使用HttpUrlConnection后请求参数作为json对象。但它不为我工作，我这样做的方式如下：

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。我做错了什么吗？如何解决这个问题。需要帮助。谢谢。
出现以下系统错误

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)

android

来源：https://stackoverflow.com/questions/16976240/httpurlconnection-with-post-request-and-parameter-as-json-object-android

2条答案

按热度按时间

luaexgnf1#

发生此错误是由于401响应代码。

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将OAuth验证器代码发送到此时预期的位置。
您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();
    }

赞(0）回复(0）举报 2023-01-03

zbwhf8kr2#

我强烈推荐使用Unirest，它是一个简单的库，用于发出http请求，它有一个非常容易使用的API，它不会使mem过载太多，这里有一个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()

赞(0）回复(0）举报 2023-01-03

我来回答

HttpUrlConnection，包含发布请求和参数，作为Json对象android

2条答案

相关问题

热门标签

最新问答