Question

我正在尝试通过HTTP POST请求的主体将数据（学生ID＃）从Android应用程序发送到PHP文件。然后PHP文件将数据（现在只是一个字符串）发送回应用程序。但是，我的php似乎无法从请求中读取我的POST数据（学生ID＃）。

我的.java文件：

        JSONObject jsonParam = new JSONObject();
        DataOutputStream printout;
        String idIN = params[0];
        jsonParam.put("id_in", idIN);
        BufferedReader input;
        String result;


        URL url = null;
        HttpURLConnection urlConnection = null;

        url = new URL("http://10.0.2.2/project/connector.php");
        urlConnection = (HttpURLConnection) url.openConnection();

        //prepare request
        urlConnection.setRequestProperty("Content-Type", "application/json");
        urlConnection.setReadTimeout(10000);
        urlConnection.setConnectTimeout(15000);  
        urlConnection.setRequestMethod("POST");
        urlConnection.setDoInput(true);
        urlConnection.setDoOutput(true);

        printout = new DataOutputStream(urlConnection.getOutputStream ());

        //printout.write(jsonParam);
        printout.writeUTF(URLEncoder.encode(jsonParam.toString(),"UTF-8"));

        //this returns: {"id_in":"1010101"}
        Log.d("json out: ", jsonParam.toString());

        printout.flush();
        printout.close();

        int response = -1;

        response = urlConnection.getResponseCode();

        input = new BufferedReader(new InputStreamReader(urlConnection.getInputStream()));
        result = input.readLine();

        // This returns 200
        Log.d("response code: ", result);

        urlConnection.disconnect();

我的.PHP文件：

 <?php

 header('Content-Type: text/html; charset=utf-8'); 

 $id_in = "";

 echo "test1";

 $id_in = trim($_POST['id_in']);

 echo "test2";

 if (isset($_POST['id_in'])) {
     echo "good";
 }else{
     echo "bad";
 }

 mysqli_close($myconn);

我的Android应用程序正在接收200响应代码并正在接收＆＃34; test1＆＃34;在响应中，但不是＆＃34; test2＆＃34;，所以当我的PHP文件尝试读取POST数据时必须出现问题：$id_in = trim($_POST['id_in']);

Answer 1

不要忘记使用urlConnection.connect（）;

从PHP应用程序发送的POST数据未被PHP文件接收

1 个答案: