如何将2个变量从JSON传递给PHP

Question

我正在创建android应用程序，以便从android设备与我的localhost服务器进行通信。我想将matrix_id和course_code传递给PHP查询。

Logcat错误： 04-02 13:30:45.060: E/JSON Parser(17463): Error parsing data [No value for exam_place]

我为此搜索了很多，但我不满意我的查询。

请帮助我。

这是PHP脚本。

   // array for JSON response
   $response = array();

include('include/db_connect.php');
mysql_select_db($database_localhost,$localhost);


   // check for post data 
  if (isset($_GET["matrix_id"]) && $_GET["course_code"]) {
  echo $matrix_id = $_GET['matrix_id'];
  echo $course_code = $_GET['course_code'];

// get a subject from subject table
$result = mysql_query("SELECT * FROM student WHERE matrix_id = '$matrix_id' AND course_code = '$course_code'");

if (!empty($result)) {
    // check for empty result
    if (mysql_num_rows($result) > 0) {

        $result = mysql_fetch_array($result);

        $student = array();
        $student["matrix_id"] = $result["matrix_id"];
        $student["course_code"] = $result["course_code"];
        $student["name"] = $result["name"];
        $student["email"] = $result["email"];
        $student["exam_place"] = $result["exam_place"];
        $student["exam_date"] = $result["exam_date"];
        $student["time"] = $result["time"];
        $student["desk_no"] = $result["desk_no"];

        // success
        $response["success"] = 1;

        // user node
        $response["student"] = array();

        array_push($response["student"], $student);

        // echoing JSON response
        echo json_encode($response);
    } else {
        // no subject found
        $response["success"] = 0;
        $response["message"] = "No subject found";

        // echo no users JSON
        echo json_encode($response);
    }
  } else {
    // no subject found
    $response["success"] = 0;
    $response["message"] = "No subject found";

    // echo no users JSON
    echo json_encode($response);
}
} else {
 // required field is missing
 $response["success"] = 0;
 $response["message"] = "Required field(s) is missing";

// echoing JSON response
echo json_encode($response);
}
 ?>

这是我的java脚本。

 protected String doInBackground(String... args) {          

        // Building Parameters
        List<NameValuePair> params = new ArrayList<NameValuePair>();

        // post matrix_id and course_code GET parameters
        params.add(new BasicNameValuePair(TAG_MATRIX_ID, matrix_id));
        params.add(new BasicNameValuePair(TAG_COURSE_CODE, course_code));

        // getting JSON string from URL
        JSONObject json = jsonParser.makeHttpRequest(url_subject_details, "GET",params);

        // Check your log cat for JSON reponse
        Log.d("Single Subject JSON: ", json.toString());

        int success;    
        try {
            // json success tag
            success = json.getInt(TAG_SUCCESS);

            if (success == 1) {

                if(json != null){   
                    // successfully received subject details
                    JSONArray subjectObj = json.getJSONArray(TAG_STUDENT); // JSON Array

                    // get first subject object from JSON Array
                    JSONObject subject = subjectObj.getJSONObject(0);

                    exam_place = subject.getString(TAG_EXAM_PLACE);
                    exam_date = subject.getString(TAG_EXAM_DATE);
                    time = subject.getString(TAG_TIME);
                    desk_no = subject.getString(TAG_DESK_NO);                               
                }           

            }
            else {
                //no subject found
            }
        }
            catch (JSONException e) {
                Log.e("JSON Parser", "Error parsing data [" + e.getMessage()+"] "+json);
            }
        return null;    
    }