Android Volley：通过用户指定的 id 从 MySQL 数据库中选择数据

Question

我正在使用 Volley 编写一个简单的 Android 应用程序。我想知道如何通过 ID 从 MySQL 数据库中选择名字和姓氏，哪个用户在应用程序中输入了 editText。这是我的 PHP 脚本：

 <?php
    include 'connection.php';

    global $connect;
    $id = $_POST["id"];

    $query = "SELECT firstName, lastName FROM users WHERE id = '$id'";

    $result = mysqli_query($connect, $query);
    $number_of_rows = mysqli_num_rows($result);

    $response = array();

    if($number_of_rows > 0) {
        while($row = mysqli_fetch_assoc($result)) {
            $response[] = $row;
        }
    }

    header('Content-Type: application/json');
    echo json_encode(array("users"=>$response));
    mysqli_close($connect);

?>

如果我在代码中指定了 ID，它会返回我要求的 JSON 格式的数据，所以脚本和数据库都可以。我得到 $id=1 的响应是：

{"users":[{"firstName":"Jonash","lastName":"Corvin"}]}

这是我的 StringRequest 代码：

    StringRequest stringRequest = new StringRequest(Request.Method.POST, url, new Response.Listener<String>() {
    @Override
    public void onResponse(String response) {
        try {
            JSONArray jsonArray = new JSONArray(response);
            JSONObject jsonObject = jsonArray.getJSONObject(0);

            String firstName = jsonObject.getString("firstName");
            String lastName = jsonObject.getString("lastName");

            firstNameTV.setText(firstName);
            lastNameTV.setText(lastName);

        } catch (JSONException e) {
            e.printStackTrace();
        }
    }
}, new Response.ErrorListener() {
    @Override
    public void onErrorResponse(VolleyError error) {
        Toast.makeText(MainActivity.this, "Something went wrong",Toast.LENGTH_LONG).show();
        error.printStackTrace();
    }
}) {
    @Override
    protected Map<String, String> getParams() throws AuthFailureError {
        Map<String,String> parameters = new HashMap<String, String>();
        parameters.put("id", idEditText.getText().toString());
        return parameters;
    }
    };
queue.add(stringRequest);

不幸的是，它没有做任何事情......它甚至没有显示任何错误或 toast 消息。你知道怎么解决吗？

Answer 1

在您的 PHP 代码中将 $connect 更改为 $conn，这将解决您的问题

<?php
    include 'connection.php';

    global $connect;
    $id = $_POST["id"];

    $query = "SELECT firstName, lastName FROM users WHERE id = '$id'";

    $result = mysqli_query($conn, $query);  //<--------change
    $number_of_rows = mysqli_num_rows($result);

    $response = array();

    if($number_of_rows > 0) {
        while($row = mysqli_fetch_assoc($result)) {
            $response[] = $row;
        }
    }

    header('Content-Type: application/json');
    echo json_encode(array("users"=>$response));
    mysqli_close($conn);  //<--------change

?>

Answer 2

     StringRequest stringRequest = new StringRequest(Request.Method.POST, URL_TRANSACTION_DETAILS,
                    new Response.Listener<String>() {
                        @Override
                        public void onResponse(String response) {
                            progressDialog.dismiss();
                            Log.i(TAG,response.toString());
                            try {
                                JSONObject jsonObject = new JSONObject(response);
                                String success = jsonObject.getString("success");
                                JSONArray jsonArray = jsonObject.getJSONArray("read");

                                if (success.equals("1")) {
                                    Toast.makeText(Activity.this,"Data Fetched!",Toast.LENGTH_SHORT).show();
                                    for (int i = 0; i < jsonArray.length(); i++) {

                                        JSONObject object = jsonArray.getJSONObject(i);

                                    String firstName= object.getString("firstName").trim();
                                    String lastName= object.getString("lastName").trim();
                                    }
}
catch (JSONException e) {
                            e.printStackTrace();
                            Toast.makeText(Activity.this,"Json Login Error " +e.toString(), Toast.LENGTH_SHORT).show();
                        }}},
new Response.ErrorListener() {
                    @Override
                    public void onErrorResponse(VolleyError error) {
                        Toast.makeText(Activity.this,"Volley Login Error " +error.toString(), Toast.LENGTH_SHORT).show();

                    }
                })

        {
            @Override
            protected Map<String, String> getParams() throws AuthFailureError {
                Map<String, String> params = new HashMap<>();
                params.put("id", id);
                return params;
            }
        };

Answer 3

  <?php
    if ($_SERVER['REQUEST_METHOD']=='POST') {
        $id = $_POST['id'];
        require_once 'connection.php'
        $sql = "SELECT firstName, lastName FROM users WHERE id='$id' ";
        $response = mysqli_query($conn, $sql);
        $result = array();
        $result['read'] = array();
        if( mysqli_num_rows($response) === 1 ) {      
            if ($row = mysqli_fetch_assoc($response)) {
                 $h['firstName']        = $row['firstName'] ;
                 $h['lastName']       = $row['lastName'] ;
                 array_push($result["read"], $h);
                 $result["success"] = "1";
                 echo json_encode($result);
            }
       }
     }else {
         $result["success"] = "0";
         $result["message"] = "Error!";
         echo json_encode($result);
         mysqli_close($conn);
     }
     ?>

Answer 4

{"users":[{"firstName":"Jonash","lastName":"Corvin"}]}

根据给定的 JSON，您需要像这样解析 JSON：

JSONObject jsonobject = new JSONObject(response);
JSONArray jsonarray = jsonobject.getJSONArray("users");
JSONObject data = jsonArray.getJSONObject(0);

String firstName = data.getString("firstName");
String lastName = data.getString("lastName");