PHP 检查数据库中的项目

Question

我是 PHP 新手，我正在尝试通过 MySQLi 连接检查 MySQL 数据库中是否存在用户。我尝试检查的用户名存储在名为 $code 的变量中。我应该在下面的语句中插入什么来让它检查 $code？

$stmt = $this->db->prepare("SELECT id_user FROM members WHERE username = ??? ($code)");  

感谢您的帮助。

编辑:这是我的代码:

class RedeemAPI {

private $db;

// Constructor - open DB connection
function __construct() {
    $this->db = new mysqli('localhost:3306', 'username', 'password', 'db');
    $this->db->autocommit(FALSE);
}

// Destructor - close DB connection
function __destruct() {
    $this->db->close();
}

// Main method to redeem a code
function redeem() {

    // Check for required parameters
    if (isset($_POST["username"])) {

        // Put parameters into local variables
        $code = $_POST["username"];

        // Look up code in database
        $id_user= 0;

        $stmt = $this->db->prepare('SELECT id_user FROM members WHERE username = ?');  

        $stmt->bind_param("s", $code);

        $stmt->execute();

        $stmt->bind_result($id_user);

        while ($stmt->fetch()) {
            break;
        }
        $stmt->close();

        // Bail if code doesn't exist
        if ($id_user <= 0) {
            sendResponse(400, 'Invalid code');

            return false;
        }

        // Return username, encoded with JSON
        $result = array("username" => $code);
        sendResponse(200, json_encode($result));
        return true;
    }
    sendResponse(400, 'Invalid request');
    return false;

}

}

Answer 1

参见 mysqli_stmt::bind_param() :

$stmt = $this->db->prepare("SELECT id_user FROM members WHERE username = ?");  
$stmt->bindParam('s', $code);
$stmt->exeucte();