Question

我正在遵循指南（https://www.youtube.com/watch?v=a3Wxi-ZvUng），现在我已经完成了登录系统的这一部分，但现在不起作用了，我做了所有工作，就像教程中的家伙一样，并且我已经看了然后三遍，以确保我的代码与他的代码相同。我有它在本地服务器上运行，并在数据库中有两个用户。（我遵循所有这些教程）

我已经在评论部分中要求解决方案，但是视频来自三年前，还没有答复。

function getLogin($conn) {
    if (isset($_POST['loginSubmit'])) {
        $uid = $_POST['uid'];
        $pwd = $_POST['pwd'];

        $sql = "SELECT * FROM user WHERE uid='$uid' AND pwd='$pwd'";
        $result = $conn->query($sql);
        if(mysqli_num_rows($result) > 0) {  
            if($row = $result->fetch_assoc()) { 
                $_SESSION['id'] = $row['id'];
                header("Location: index.php?loginsuccess");
                exit();
            }
        }
        else {
            header("Location: include.php?loginfailed");
            exit();
        }   
    }
}

function userLogout() {
    if (isset($_POST['logoutSubmit'])) {
        session_start();
        session_destroy();
        header("Location: include.php");
        exit();
    }   
}

查看

<?php
    echo "<form method='POST' action='".getLogin($conn)."'>
        <input type='text' name='iud'>
        <input type='password' name='pwd'>
        <button type='submit' name='loginSubmit'>Login</button>
    </form>";
    echo "<form method='POST' action='".userLogout()."'>
        <button type='submit' name='logoutSubmit'>Logout</button>
    </form>";

    if (isset($_SESSION['id'])) {
        echo "You are logged in!";
    } else {
        echo "You are not logged in!";
    }

?>

Answer 1

尝试一下：

<?php
if (isset($_POST['logoutSubmit'])) {
    session_start();
    session_destroy();
    header("Location: include.php");
    exit();
}   
 if (isset($_POST['loginSubmit'])) {
    $uid = $_POST['uid'];
    $pwd = $_POST['pwd'];

    $sql = "SELECT * FROM user WHERE uid='$uid' AND pwd='$pwd'";
    $result = $conn->query($sql);
    if(mysqli_num_rows($result) > 0) {  
        if($row = $result->fetch_assoc()) { 
            $_SESSION['id'] = $row['id'];
            header("Location: index.php?loginsuccess");
            exit();
        }
    }
    else {
        header("Location: include.php?loginfailed");
        exit();
    }   
} ?>

<form method='POST' action="#">
    <input type='text' name='iud'>
    <input type='password' name='pwd'>
    <button type='submit' name='loginSubmit'>Login</button>
    <button type='submit' name='logoutSubmit'>Logout</button>
</form>

请记住，这具有SQL注入和纯文本密码漏洞，因此您不应将其实时发布到互联网上。

登录系统一直说我无法登录

1 个答案: