我編寫了這段代碼，它應該將一堆值插入到數據庫中，但是每當我運行此代碼時，它都會顯示“致命錯誤：未捕獲的異?！甈DOException’，消息為‘SQLSTATE[42000]：語法錯誤或訪問沖突：1064您的 SQL 語法有錯誤；請檢查與您的 MySQL 服務器版本相對應的手冊，以了解要使用的正確語法“附近”和“PDOException：SQLSTATE[42000]：語法錯誤或訪問沖突：1064 您的 SQL 有錯誤語法；檢查與您的 MySQL 服務器版本相對應的手冊，以獲取在附近使用的正確語法。有人可以幫忙嗎？我的代碼，根據這些錯誤消息，錯誤在 INSERT 部分的某處：mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);if (session_status() == PHP_SESSION_NONE) {  session_start();}function getValue1() {    if (!isset($_GET[r])) {        return false;        $referral_code = "-";    }    return $_GET[r];    $referral_code = $_GET[r];}// initializing variables$username = "";$email    = "";$errors = array();    require("php_db_info.php");   $dsn = 'mysql:host=127.0.0.1;dbname=user_db;charset=utf8';       $conn = new PDO($dsn, $username1, $password);       $conn->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);       $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);// REGISTER USERif (isset($_POST['reg_user'])) {  // receive all input values from the form  $username = $_POST['username'];  $email = $_POST['email'];  $password_1 = $_POST['password_1'];  $password_2 = $_POST['password_2'];  if (getValue('r')) {    $referral_code = $_GET['r'];} else {      $referral_code = "-";  }   if (empty($username)) { array_push($errors, "Username is required."); }  if (empty($email)) { array_push($errors, "Email is required."); }  if (empty($password_1)) { array_push($errors, "Password is required."); }  if ($password_1 != $password_2) {      array_push($errors, "The two passwords do not match.");  }  if(preg_match("/[^a-zA-Z0-9_-]/i", $username)) {    array_push($errors, "You can only use letters, numbers, underscores and dashes.");  }  if (strlen($username) <= 1) {    array_push($errors, "Your username must have at least 2 characters.");  }   if (strlen($username) >= 22) {  array_push($errors, "Your username must have below 22 characters.");  }   require "server_bannedwords.php";