Question

我不明白为什么我无法将登录的用户名与会话相关联。如果你能帮助我，我会非常感激吗？

以下是示例代码：

session_start();

include 'includes/connect.php';

if (isset($_POST['user'], $_POST['pass'])) {
    //assigning parameters to variables.
    $user = $_POST['user'];
    $pass = $_POST['pass'];

    $sql = 'SELECT id FROM users WHERE username=(:user) AND password=(:pass)';

    try {
        $connect = new PDO("mysql:host=$host; dbname=$dbname;", $username, $password);
        $connect->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
        $values = $connect->prepare($sql);
        $values->bindParam('user', $user, PDO::PARAM_STR);
        $values->bindParam('pass', $pass, PDO::PARAM_STR);
        $values->execute();
        $row_count = $values->rowCount();
        $row = $values->fetch(PDO::FETCH_ASSOC);

        ////////////////////////////
        //user and pass section [DB]
        //$usr_db = $row['username'];
        //$pwd_db = $row['password'];
        ////////////////////////////

        if ($row_count === 1) {
            $_SESSION['Logged_in'] = $user;
            echo 'THIS TEXT REMAINS..';
        }
    } catch (PDOException $e) {
        echo 'Error: ' . $e->getMessage();
    }
}

Answer 1

您还没有给我们足够的信息来帮助您，但我会做出有根据的猜测。你想要文本＆＃34;这个文字仍然存在。＆＃34;一直出现。为什么这不会发生是因为除非在每个请求上发布用户名和密码，否则执行永远不会到达那一点，我认为这不是发生的事情。如果您重新组织这样的代码，您可能会得到您期望的结果。

session_start();

include 'includes/connect.php';

if (isset($_POST['user'], $_POST['pass'])) {
    //assigning parameters to variables.
    $user = $_POST['user'];
    $pass = $_POST['pass'];

    $sql = 'SELECT id FROM users WHERE username=(:user) AND password=(:pass)';

    try {
        $connect = new PDO("mysql:host=$host; dbname=$dbname;", $username, $password);
        $connect->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
        $values = $connect->prepare($sql);
        $values->bindParam('user', $user, PDO::PARAM_STR);
        $values->bindParam('pass', $pass, PDO::PARAM_STR);
        $values->execute();
        $row_count = $values->rowCount();
        $row = $values->fetch(PDO::FETCH_ASSOC);

        ////////////////////////////
        //user and pass section [DB]
        //$usr_db = $row['username'];
        //$pwd_db = $row['password'];
        ////////////////////////////

        if ($row_count === 1) {
            $_SESSION['Logged_in'] = $user;
        }
    } catch (PDOException $e) {
        echo 'Error: ' . $e->getMessage();
    }
}

// Moved the login check down here so it's executed on every request.
if (array_key_exists('Logged_in', $_SESSION) && $_SESSION['Logged_in'] !== null) {
    echo 'THIS TEXT REMAINS..';
}

会话未与用户名PHP关联

1 个答案: