在Android中使用POST方法发送参数

Question

我想使用POST方法向服务器发送参数。

我使用ServiceHandler.java文件来处理请求

public class ServiceHandler { 

    static String response = null; 
    public final static int GET = 1; 
    public final static int POST = 2; 

    public ServiceHandler() { 

    } 

    /** 
    * Making service call 
    * @url - url to make request 
    * @method - http request method 
    * */ 
    public String makeServiceCall(String url, int method) { 
     return this.makeServiceCall(url, method, null); 
    } 

    /** 
    * Making service call 
    * @url - url to make request 
    * @method - http request method 
    * @params - http request params 
    * */ 
    public String makeServiceCall(String url, int method, 
            List<NameValuePair> params) { 
     try { 
      // http client 
      DefaultHttpClient httpClient = new DefaultHttpClient(); 
      HttpEntity httpEntity = null; 
      HttpResponse httpResponse = null; 

      // Checking http request method type 
      if (method == POST) { 
       HttpPost httpPost = new HttpPost(url); 

       // adding post params 
       if (params != null) { 
        httpPost.setEntity(new UrlEncodedFormEntity(params)); 
       } 

       httpResponse = httpClient.execute(httpPost); 

      } else if (method == GET) { 
       // appending params to url 
       if (params != null) { 
        String paramString = URLEncodedUtils 
          .format(params, "utf-8"); 
        url += "?" + paramString; 
       } 
       HttpGet httpGet = new HttpGet(url); 

       httpResponse = httpClient.execute(httpGet); 

      } 
      httpEntity = httpResponse.getEntity(); 
      response = EntityUtils.toString(httpEntity); 

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

     return response; 

    } 
} 

而且我从我的活动文件传递参数

// BACKGROUND EXECUTION STARTS 
     @Override 
     protected Void doInBackground(Void... arg0) { 


      // Creating service handler class instance 
      ServiceHandler sh = new ServiceHandler(); 

      List<NameValuePair> params = new ArrayList<NameValuePair>(); 
      params.add(new BasicNameValuePair("pkey", "pvalue")); 
      sh.makeServiceCall(url, ServiceHandler.POST, params); 


      // Making a request to url and getting response 
      String jsonStr = sh.makeServiceCall(url, ServiceHandler.POST); 

      Log.d("Response: ", "> " + jsonStr); 

我能够连接到我的服务器，但它没有从我的应用程序接收任何参数。

有人可以修改它，并发送工作代码 谢谢

Answer 1

尝试使用此代码：

ServiceHandler sh = new ServiceHandler(); 
List<NameValuePair> params = new ArrayList<NameValuePair>(); 

@Override 
protected void onPreExecute() { 
    // TODO Auto-generated method stub 
    super.onPreExecute(); 


    params.add(new BasicNameValuePair("pkey", "pvalue")); 

} 

@Override 
protected String doInBackground(String... param) { 
     // TODO Auto-generated method stub 
     String json; 
     try { 
      json = sh.makeServiceCall(url, ServiceHandler.POST, params); 

      Log.d("Response: ", "> " + json); 

     } catch (Exception e) { 
      // TODO Auto-generated catch block 
      e.printStackTrace(); 
     } 

     return null; 
    } 

@Override 
protected void onPostExecute(String result) { 
    // TODO Auto-generated method stub 
    super.onPostExecute(result); 


    } 

Answer 2

您应HttpURLConnection类来代替：

public class MessageSender { 
    private int responseCode; 


    public boolean sendPost(YourParameter param) { 

     HttpURLConnection connection; 
     try { 
      URL gcmAPI = new URL("your api url"); 
      connection = (HttpURLConnection) gcmAPI.openConnection(); 

      connection.setRequestMethod("POST"); 
      connection.setRequestProperty("Content-Type", "application/json"); 
          connection.setDoOutput(true); 

      ObjectMapper mapper = new ObjectMapper(); 
      mapper.setVisibility(PropertyAccessor.FIELD, JsonAutoDetect.Visibility.ANY); 
      DataOutputStream dataOutputStream = new DataOutputStream(connection.getOutputStream()); 

      mapper.writeValue(dataOutputStream, param); 

      dataOutputStream.flush(); 
      dataOutputStream.close(); 

      responseCode = connection.getResponseCode(); 
     } catch (IOException e) { 
      e.printStackTrace(); 
     } 
     if (responseCode == 200) { 
      Log.i("Request Status", "This is success response status from server: " + responseCode); 
      return true; 
     } else { 
      Log.i("Request Status", "This is failure response status from server: " + responseCode); 
      return false; 
     } 
    } 
} 

在你的活动：

MessageSender mgsSender = new MessageSender(); 
        new AsyncTask<Void, Void, Void>() { 
         @Override 
         protected Void doInBackground(Void... params) { 

          mgsSender.sendPost(mgsContent); 

          } 
          return null; 
         } 
        }.execute(); 

Answer 3

/*更改您的服务处理类此行*/

public JSONObject makeHttpRequest(String url, String method, 
List<NameValuePair> params) { 
if(method == "POST"){ 
}else if(method == "GET"){ 
} 
} 

/*论doinbackground乌尔活动课*/

sh.makeServiceCall(url,"POST", params);