我编写了这段代码，它应该将一堆值插入到数据库中，但是每当我运行此代码时，它都会显示“致命错误：未捕获的异常‘PDOException’，消息为‘SQLSTATE[42000]：语法错误或访问冲突：1064您的 SQL 语法有错误；请检查与您的 MySQL 服务器版本相对应的手册，以了解要使用的正确语法“附近”和“PDOException：SQLSTATE[42000]：语法错误或访问冲突：1064 您的 SQL 有错误语法；检查与您的 MySQL 服务器版本相对应的手册，以获取在附近使用的正确语法。有人可以帮忙吗？
我的代码，根据这些错误消息，错误在 INSERT 部分的某处：
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);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
// REGISTER USER
if (isset($_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";