将参数发送到php并以JSON格式获取结果

我有将参数发送到php并以JSON获取结果的代码：

private void details_location(int id){

        HashMap<String, String> parameter = new HashMap<>();
        parameter.put("id", id);

        JsonObjectRequest jsonObjReq = new JsonObjectRequest(Method.GET, url, new JSONObject(parameter), new Response.Listener<JSONObject>() {
            @Override
            public void onResponse(JSONObject response) {
                Log.d(TAG, response.toString());

                try {
                    JSONArray respon=(JSONArray)response.get("location");

                    for (int i = 0; i < respon.length(); i++) {
                        JSONObject person = (JSONObject) respon.get(i);
                        id_location= person.getString("id_location");
                        address= person.getString("address");                   
                    }
                } catch (JSONException e) {
                    e.printStackTrace();
                    Toast.makeText(getApplicationContext(),
                            "Error: " + e.getMessage(),
                            Toast.LENGTH_LONG).show();
                }
            }
        }, new Response.ErrorListener() {

            @Override
            public void onErrorResponse(VolleyError error) {
                VolleyLog.d(TAG, "Error: " + error.getMessage());
                Toast.makeText(getApplicationContext(),
                        error.getMessage(), Toast.LENGTH_SHORT).show();
            }
        });
    }

<?php
    require_once('db_connect.php');
    if(isset($_GET['id'])){
        $id=$_GET['id'];
        $action="SELECT * FROM locaions WHERE id_location=\"$id\";";
        $query=mysqli_query($db_connect, $action);

        if (mysqli_num_rows($query) > 0)
        {
            $json['location']=array();
            while($row=mysqli_fetch_assoc($query)){
                $data=array();
                $data["id_location"]=$row["id_location"];
                $data["address"]=$row["address"];

                array_push($json['location'], $data);
            }
        }
        mysqli_close($db_connect);
        echo json_encode($json);
    }
    else{
        $action="SELECT * FROM locaions WHERE id_location=4;";
        $query=mysqli_query($db_connect, $action);

        if (mysqli_num_rows($query) > 0)
        {
            $json['location']=array();
            while($row=mysqli_fetch_assoc($query)){
                $data=array();
                $data["id_location"]=$row["id_location"];
                $data["address"]=$row["address"];

                array_push($json['location'], $data);
            }
        }
        mysqli_close($db_connect);
        echo json_encode($json);
    }   
?>

private void details\u位置（int id）{
HashMap参数=新的HashMap（）；
参数put（“id”，id）；
JsonObjectRequest JSONObjectReq=新的JsonObjectRequest（Method.GET、url、新的JSONObject（参数）、新的Response.Listener（）{
@凌驾
公共void onResponse（JSONObject响应）{
Log.d（TAG，response.toString（））；
试一试{
JSONArray respon=（JSONArray）response.get（“位置”）；
对于（int i=0；i

我有处理查询Mysql的php文件，并有JSON输出：

private void details_location(int id){

        HashMap<String, String> parameter = new HashMap<>();
        parameter.put("id", id);

        JsonObjectRequest jsonObjReq = new JsonObjectRequest(Method.GET, url, new JSONObject(parameter), new Response.Listener<JSONObject>() {
            @Override
            public void onResponse(JSONObject response) {
                Log.d(TAG, response.toString());

                try {
                    JSONArray respon=(JSONArray)response.get("location");

                    for (int i = 0; i < respon.length(); i++) {
                        JSONObject person = (JSONObject) respon.get(i);
                        id_location= person.getString("id_location");
                        address= person.getString("address");                   
                    }
                } catch (JSONException e) {
                    e.printStackTrace();
                    Toast.makeText(getApplicationContext(),
                            "Error: " + e.getMessage(),
                            Toast.LENGTH_LONG).show();
                }
            }
        }, new Response.ErrorListener() {

            @Override
            public void onErrorResponse(VolleyError error) {
                VolleyLog.d(TAG, "Error: " + error.getMessage());
                Toast.makeText(getApplicationContext(),
                        error.getMessage(), Toast.LENGTH_SHORT).show();
            }
        });
    }

<?php
    require_once('db_connect.php');
    if(isset($_GET['id'])){
        $id=$_GET['id'];
        $action="SELECT * FROM locaions WHERE id_location=\"$id\";";
        $query=mysqli_query($db_connect, $action);

        if (mysqli_num_rows($query) > 0)
        {
            $json['location']=array();
            while($row=mysqli_fetch_assoc($query)){
                $data=array();
                $data["id_location"]=$row["id_location"];
                $data["address"]=$row["address"];

                array_push($json['location'], $data);
            }
        }
        mysqli_close($db_connect);
        echo json_encode($json);
    }
    else{
        $action="SELECT * FROM locaions WHERE id_location=4;";
        $query=mysqli_query($db_connect, $action);

        if (mysqli_num_rows($query) > 0)
        {
            $json['location']=array();
            while($row=mysqli_fetch_assoc($query)){
                $data=array();
                $data["id_location"]=$row["id_location"];
                $data["address"]=$row["address"];

                array_push($json['location'], $data);
            }
        }
        mysqli_close($db_connect);
        echo json_encode($json);
    }   
?>

---

当我运行程序时，结果总是以

id\u location=4

（语句else）显示位置。为什么

isset（$\u GET['id']）=false

？如何将参数发送到PHP并在JSON中获得结果？

这是将参数发送到ur PHP服务器的代码

protected String doInBackground(String... params) {
            // TODO Auto-generated method stub

            try {
                String parameter1 = (String) params[0];
                String parameter2 = (String) params[1];

                String link = "Mention ur url link here/parameter1+parameter1 +parameter2+parameter2+";

                URL url = new URL(link);
                HttpClient client = new DefaultHttpClient();
                HttpGet request = new HttpGet();
                request.setURI(new URI(link));
                HttpResponse response = client.execute(request);
                BufferedReader in = new BufferedReader(new InputStreamReader(
                        response.getEntity().getContent()));

                StringBuffer sb = new StringBuffer("");
                String line = "";

                while ((line = in.readLine()) != null) {
                    sb.append(line);
                    break;
                }
                in.close();
                return sb.toString();
            }

            catch (Exception e) {
                return new String("Exception: " + e.getMessage());
            }
        }

在执行后，php应该返回json格式的数据，您可以在执行后处理这些数据

要回答“为什么isset（$\u GET['id']）为false”的问题，您应该打印您收到的值以及您试图访问的URL，这样您就可以更好地了解发生了什么。