Mysql插入返回没有回到ajax

Question

我正在尝试将数据插入 MySQL 数据库。数据插入成功。但在成功插入ajax后我没有得到返回消息。我也使用最新的 jquery 版本和 PHP 版本。

剧本

<script>
  $(document).ready(function () {
    $('#data_form').on("submit",() => {
      var formData = $('#data_form').serialize();
      if (!$.active) {
        $.ajax({
          type: 'POST',
          url: '../../../controller/add_user.php',
          data :formData,
          cache: false,
          success: function (data) {
            console.log(data.message);
          },
          error: function(xhr, status, error) {
            console.log(xhr.responseJSON.message);
          }
        });
      }
    });
  });
</script> 

add_user.php

<?php
include_once('../model/database_11.php');
$db_obj = new Dbconnection();
$dbconn = $db_obj->ConnectToDB();

class admin
{
    public $dbconn;
    public $name;
    public $account_type;
    public $telephone = null;
    public $mobile;
    public $address;
    public $pincode;
    public $email_user;
    public $gender;
    public $dob;
    public $status;
    public $username;
    public $password;
    public $query;

    public function __construct($dbconn, $name, $account_type, $telephone, $mobile, $address, $pincode, $email_user, $gender, $dob, $status, $username, $password)
    {
        $this->dbconn = $dbconn;
        $this->name = $name;
        $this->account_type = $account_type;
        $this->telephone = $telephone;
        $this->mobile = $mobile;
        $this->address = $address;
        $this->pincode = $pincode;
        $this->email_user = $email_user;
        $this->gender = $gender;
        $this->dob = date('Y-m-d', strtotime($dob));
        $this->status = $status;
        $this->username = $username;
        $this->password = $password;
    }
    function adduser()
    {
        $this->query = "INSERT INTO `users` (`ID_USER`, `WEB_USERNAME`, `WEB_USERPASSWORD`, `USER_TYPE`, `NAME`, `IMAGE`, `TELEPHONE`, `MOBILE`, `ADDRESS`, `PIN`, `EMAIL`, `GENDER`, `DATE_OF_BIRTH`, `STATUS`) VALUES (NULL, '$this->username', '$this->password', '$this->account_type', '$this->name', NULL, '$this->telephone', '$this->mobile', '$this->address', '$this->pincode', '$this->email_user', '$this->gender', '$this->dob', '$this->status');";
        if ($this->dbconn->query($this->query) === TRUE) {
            return "200";
        } else {
            return "500";
        }
    }
    public function __destruct()
    {
        $this->dbconn->close();
    }

}
extract($_POST);
$admin = new admin($dbconn, $name, $account_type, $telephone, $mobile, $address, $pincode, $email_user, $gender, $dob, $status, $username, $password);
$res = $admin->adduser();
echo $res;
?>

我尝试将 echo 语句更改为 http_response_code。但还是不行。

Answer 1

正如您在评论中看到的，您需要发回 json

function adduser()
{
    $response = array();

    $this->query = "INSERT INTO `users` (`ID_USER`, `WEB_USERNAME`, `WEB_USERPASSWORD`, `USER_TYPE`, `NAME`, `IMAGE`, `TELEPHONE`, `MOBILE`, `ADDRESS`, `PIN`, `EMAIL`, `GENDER`, `DATE_OF_BIRTH`, `STATUS`) VALUES (NULL, '$this->username', '$this->password', '$this->account_type', '$this->name', NULL, '$this->telephone', '$this->mobile', '$this->address', '$this->pincode', '$this->email_user', '$this->gender', '$this->dob', '$this->status');";

    if ($this->dbconn->query($this->query) === TRUE) {
        $response['status'] = "success";
        $response['message'] = "User added successfully";
    } else {
        $response['status'] = "error";
        $response['message'] = "Error adding user: " . $this->dbconn->error;
    }

    return json_encode($response);
}