使用StringRequest传递参数以在Volley库中获取JSON数组？

Question

我想通过将参数传递给服务器并获取JSON数组来在sql中触发自定义查询。我使用Volley库来处理html请求。但是，使用以下代码我的应用程序冻结。我通过互联网搜索过，但我无法恢复。这是代码：

StringRequest postRequest = new StringRequest(Request.Method.POST, Config.serv_url, /*url to server*/ new Response.Listener<String>()
    {
        @Override
        public void onResponse(String response) {
            try {
                JSONArray jsonArray = new JSONArray(response);

                int count = 0;
                while(count<response.length()) {
                    try {
                        JSONObject jsonObject = jsonArray.getJSONObject(count);
                                arrayList.add(new Album(jsonObject.getString("id")));
                        count++;
                    } catch (JSONException e) {
                        e.printStackTrace();
                    }
                }
            } catch (JSONException e) {
                e.printStackTrace();
            }
            adapter = new customImagesSwipe(arrayList,DisplayActivity.this);
            viewPager.setAdapter(adapter);
        }
    }, new Response.ErrorListener(){
        @Override
        public void onErrorResponse(VolleyError error) {
            Toast.makeText(DisplayActivity.this, "Error :- " + error, Toast.LENGTH_LONG).show();
        }
    }) {
        // here is params will add to your url using post method
        @Override
        protected Map<String, String> getParams() {
            Map<String, String> params = new HashMap<String, String>();
            params.put("num", image); //value for param
            //params.put("2ndParamName","valueoF2ndParam");
            return params;
        }
    };
    MySingleton.getMinstance(DisplayActivity.this).addToRequestQueue(postRequest);

这是PHP脚本：

<?php
$host = "localhost";
$user_name = "root";
$password = "";
$db_name = "imagesfromserver";
$con = mysqli_connect($host,$user_name,$password,$db_name);

$num = $_POST["num"];
$sql = "select * from album_info where (lower(id) LIKE 'num%');";
$result = mysqli_query($con,$sql);
$response = array();

while($row = mysqli_fetch_array($result))
{
    array_push($response,array("id"=>$row["id"],"title"=>$row["title"]));
}

echo json_encode($response);

mysqli_close($con);

?>

我没有在客户端服务器上工作的先验知识。我是从互联网上搜索实现的。

更新

JsonArrayRequest jsonArrayRequest = new JsonArrayRequest(Request.Method.POST, Config.serv_url, new Response.Listener<JSONArray>() {
        @Override
        public void onResponse(JSONArray response) {

            int count = 0;
            while(count<response.length()){
                try {
                    JSONObject jsonObject = response.getJSONObject(count);
                    arrayList.add(new Album(jsonObject.getString("id"), jsonObject.getString("title")));
                    count++;
                } catch (JSONException e) {
                    e.printStackTrace();
                }
            }

            adapter = new customImagesSwipe(arrayList,DisplayActivity.this);
            viewPager.setAdapter(adapter);
        }
    }, new Response.ErrorListener() {
        @Override
        public void onErrorResponse(VolleyError error) {
            Toast.makeText(DisplayActivity.this,"NULL\n" + error,Toast.LENGTH_LONG ).show();
        }
    });
    MySingleton.getMinstance(DisplayActivity.this).addToRequestQueue(jsonArrayRequest);