Question

我正在使用下面的代码发送一个http POST请求，该请求将对象发送到WCF服务。这工作正常，但如果我的WCF服务还需要其他参数，会发生什么？如何从我的Android客户端发送它们？

这是我到目前为止编写的代码：

StringBuilder sb = new StringBuilder();  

String http = "http://android.schoolportal.gr/Service.svc/SaveValues";  


HttpURLConnection urlConnection=null;  
try {  
    URL url = new URL(http);  
    urlConnection = (HttpURLConnection) url.openConnection();
    urlConnection.setDoOutput(true);   
    urlConnection.setRequestMethod("POST");  
    urlConnection.setUseCaches(false);  
    urlConnection.setConnectTimeout(10000);  
    urlConnection.setReadTimeout(10000);  
    urlConnection.setRequestProperty("Content-Type","application/json");   

    urlConnection.setRequestProperty("Host", "android.schoolportal.gr");
    urlConnection.connect();  

    //Create JSONObject here
    JSONObject jsonParam = new JSONObject();
    jsonParam.put("ID", "25");
    jsonParam.put("description", "Real");
    jsonParam.put("enable", "true");
    OutputStreamWriter out = new   OutputStreamWriter(urlConnection.getOutputStream());
    out.write(jsonParam.toString());
    out.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();  
}

Answer 1

发布参数使用POST： -

URL url;
URLConnection urlConn;
DataOutputStream printout;
DataInputStream  input;
url = new URL (getCodeBase().toString() + "env.tcgi");
urlConn = url.openConnection();
urlConn.setDoInput (true);
urlConn.setDoOutput (true);
urlConn.setUseCaches (false);
urlConn.setRequestProperty("Content-Type","application/json");   
urlConn.setRequestProperty("Host", "android.schoolportal.gr");
urlConn.connect();  
//Create JSONObject here
JSONObject jsonParam = new JSONObject();
jsonParam.put("ID", "25");
jsonParam.put("description", "Real");
jsonParam.put("enable", "true");

您错过的部分在以下内容中......即，如下..

// Send POST output.
printout = new DataOutputStream(urlConn.getOutputStream ());
printout.writeBytes(URLEncoder.encode(jsonParam.toString(),"UTF-8"));
printout.flush ();
printout.close ();

剩下的事情你可以做到。

Answer 2

尝试一些像打击一样的事情：

SString otherParametersUrServiceNeed =  "Company=acompany&Lng=test&MainPeriod=test&UserID=123&CourseDate=8:10:10";
String request = "http://android.schoolportal.gr/Service.svc/SaveValues";

URL url = new URL(request); 
HttpURLConnection connection = (HttpURLConnection) url.openConnection();   
connection.setDoOutput(true);
connection.setDoInput(true);
connection.setInstanceFollowRedirects(false); 
connection.setRequestMethod("POST"); 
connection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded"); 
connection.setRequestProperty("charset", "utf-8");
connection.setRequestProperty("Content-Length", "" + Integer.toString(otherParametersUrServiceNeed.getBytes().length));
connection.setUseCaches (false);

DataOutputStream wr = new DataOutputStream(connection.getOutputStream ());
wr.writeBytes(otherParametersUrServiceNeed);

   JSONObject jsonParam = new JSONObject();
jsonParam.put("ID", "25");
jsonParam.put("description", "Real");
jsonParam.put("enable", "true");

wr.writeBytes(jsonParam.toString());

wr.flush();
wr.close();

参考文献：

从Android发送JSON HTTP POST请求

2 个答案: