Question

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我正在编写预备语句，我收到此错误消息：

注意：未定义的索引：第34行的C：\ Program Files（x86）\ EasyPHP-12.1 \ www \ inlogg_och_lagar \ core.inc.php中的用户名

致命错误：带有消息'SQLSTATE [42000]的未捕获异常'PDOException'：语法错误或访问冲突：1064 SQL语法中有错误;查看与您的MySQL服务器版本相对应的手册，以便在C：\ Program Files（x86）\ EasyPHP-12.1 \ www \ inlogg_och_lagar \ core.inc.php中的第1行'''附近使用正确的语法：34 Stack跟踪：＃0 C：\ Program Files（x86）\ EasyPHP-12.1 \ www \ inlogg_och_lagar \ core.inc.php（34）：PDOStatement-＆gt; execute（Array）＃1 C：\ Program Files（x86）\ EasyPHP -12.1 \ www \ inlogg_och_lagar \ index.php（23）：getuserrow（'username'）＃2 {main}抛出C：\ Program Files（x86）\ EasyPHP-12.1 \ www \ inlogg_och_lagar \ core.inc.php on第34行

有人能告诉我我做错了什么吗？这是一个简单的登录脚本。

这是loginform.inc.php

    <?php 
        if (isset($_POST['username'])&&isset($_POST['password'])) {

        $username = $_POST['username'] ;    
        $password = $_POST['password'] ;

        $password_hash = md5 ($password);   


        if ($username && $password)
        {     
          $query  ="SELECT * FROM USERDATA where `username` = :username AND `password` = :password";
$stmt = $pdo->prepare($query);
$stmt->execute(array(
   ':username' => $_POST['username'],
   ':password' => $password_hash
));

$row = (bool) $stmt->fetch();
            if (!$rows)
            {   
                  echo '<p class="warning">Fel användarnamn/lösenords kombination.</p>';
            } else {    
                $_SESSION['user_id'] = $row;
                header('Location: index.php');
                exit;
            }
        } else {    
              echo '<p class="warning">Du måste fylla i ett användarnamn och lösenord</p>';
            }
    }
    ?>

        <!--- LOGIN FORM --->
        <div id="form-column">
            <p>You need to be loggedin.</p>
            <p>Fill out username and password.</p>
            <form action="<?php echo $current_file; ?>" method="POST"> 
                Användarnamn: <input type="text" name="username"> 
                Lösenord: <input type="password" name="password">
                <input type="submit" value="Log in">
            </form>
        </div>

这是core.inc.php

    <?php 
    ob_start();
    session_start();
    $current_file = $_SERVER['SCRIPT_NAME'];

    if (isset($_SERVER['HTTP_REFERER'])&&!empty($_SERVER['HTTP_REFERER'])) {
        $http_referer = $_SERVER['HTTP_REFERER'];
    }


    function loggedin () {
        if (isset($_SESSION['user_id'])&&!empty($_SESSION['user_id'])) {
            return true;
        }   else {
            return false;
        }
    }

    function getuserrow ($row) {
    $sql  ="SELECT * FROM USERDATA where id = ?".$_SESSION['user_id']."'";
    global $pdo;
    $stmt = $pdo->prepare($sql);
    $stmt->execute(array($_POST['username']));
    $row = $stmt->fetch();
}

    ?>

这是connect.inc.php

<?php

$dsn = "mysql:host=localhost;dbname=users;charset=utf8";
$opt = array(
    PDO::ATTR_ERRMODE            => PDO::ERRMODE_EXCEPTION,
    PDO::ATTR_DEFAULT_FETCH_MODE => PDO::FETCH_ASSOC
);
$pdo = new PDO($dsn,'root','', $opt);

?>

Answer 1

您的PDO对象定义为$pdo，但后来被称为$dbh：

来自connect.inc.php：
```
$pdo = new PDO($dsn,'root','', $opt);
```

来自loginform.inc.php：

global $dbh;
// ...
$stmt = $dbh->prepare($sql);

此外，loginform.inc.php似乎不包含connect.inc.php。

Answer 2

您在$stmt->execute(array($_POST['username']))错过了第二个参数它应该是

$sql  ="SELECT * FROM USERDATA where username = ? AND password = ?";
$stmt = $pdo->prepare($sql);
$stmt->execute(array($_POST['username'], md5($_POST['password'])));//<== here

Answer 3

替换它，

global $pdo;  // <- You don't need this one
$sql  ="SELECT * FROM USERDATA where username = ? AND password = ?";
$stmt = $pdo->prepare($sql);
$stmt->execute(array($_POST['username']));
$row = $stmt->fetch();

与

$query  ="SELECT * FROM USERDATA where `username` = :username AND `password` = :password";
$stmt = $pdo->prepare($query);
$stmt->execute(array(
   ':username' => $_POST['username'],
   ':password' => $password_hash
));

$row = (bool) $stmt->fetch();

为了避免像这样的情况，你最好坚持使用命名占位符而不是未命名的占位符（如?）。

准备好的声明中的致命错误

3 个答案: