Question

这是checklogin.php页面。整个想法是，根据您的状态（1或0），程序应引导您到正确的页面（red_form或yellow_form）。目前，无论我是谁（不在数据库中），这段代码都会让我登录，或者让我以数据库的身份登录，但不能正确引导我。我究竟做错了什么？

ol

Answer 1

你在错误的地方给了“其他”条件。您的登录工作正常，您可以按照代码

<?php
include("connection.php");
session_start();

?>
<!doctype html>
<html>
<head>
    <meta charset="utf-8">
    <title>Untitled Document</title>
</head>
<body>
    <?php
        if(isset($_POST['submit'])){
            $username= $_POST['username'];
            $password= sha1($_POST['password']);

            $sql = "SELECT * FROM chklogin WHERE username = '$username' AND password 
='$password'";
            $result = mysqli_query($conn, $sql);
            if($result){
                echo "Yippie";

            } else {
                echo "Error";
            }

            $rowcount = mysqli_num_rows($result); 
            echo ($rowcount);

            if($rowcount > 0)
            {
                echo "Uspw ok";
                $row = mysqli_fetch_assoc($result);
                $_SESSION['username'] = $row['username'];
                $_SESSION['password'] = $row['password'];
                $_SESSION['id'] = $row['id'];
                $_SESSION['status'] = $row['status'];
                $_SESSION['username'] = $username;

                $_SESSION['login'] = true;

                echo $_SESSION['username'];
                echo $_SESSION['id'];

                if($_SESSION['status'] == "1"){
                    header('Location: red_form.php');
                }  
            }
            else
            {
                header('Location: yellow_form.php');
            }
        }
    ?>

      <form method="post" action="">
    <input type="text" name="username" placeholder="enter name"><br><br>
    <input type="text" name="password" placeholder="enter password"><br><br>
    <input type="submit" name="submit" value="submit">
</form>
</body> 
</html>

没有错误消息，但我的登录不起作用

1 个答案: