从Android发送JSON HTTP POST请求

  • Z时代
  • 2024-01-10
  • 分类：问答

我正在使用下面的代码发送http POST请求，该请求将对象发送到WCF服务。可以，但是如果我的WCF服务还需要其他参数怎么办？如何从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();  
}

回答：

使用POST发布参数：

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");

你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 ();

其余的事情你都可以做到。

以上是 从Android发送JSON HTTP POST请求 的全部内容， 来源链接： utcz.com/qa/432925.html

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