这段代码有什么问题[PHP Cookies]

我有这个代码，但当我尝试加载页面时，它是空白的：（我用USERCOOKIEID和PASSCOOKIEID替换了cookie的实际名称，并删除了用户登录时发生的代码）

}

谢谢

看起来应该像

if(isset($_COOKIE['USERCOOKIEID']))
{ 
    $user = $_COOKIE['USERCOOKIEID']; 
    $pass = $_COOKIE['PASSCOOKIEID'];
    $check = mysql_query("SELECT * FROM `users` WHERE `username`='$user'") or die();
    if (mysql_result($check, 0, 'passwordcolnum') == $pass) {
    } else {
        //This is were the code goes for a user that is signed on
    }
} else { //what happens if they don't have the cookie
    header("Location: login.php");
}

---

另外，与其使用

mysql\u fetch\u array

，不如使用

mysql\u result

，因为肯定只有一条记录

，使用带有基本语法突出显示的IDE可以防止这些类型的错误。因此，此代码在当前状态下非常不安全，始终清理来自用户的任何数据，为了安全体系结构，任何系统都不应存储明文密码。始终不可逆地加密或散列它们，最好使用足够的salt，包括将散列与用户名、每用户salt和系统范围salt组合。这些步骤毫无用处。

if(isset($_COOKIE['USERCOOKIEID']))
{ 
    $user = $_COOKIE['USERCOOKIEID']; 
    $pass = $_COOKIE['PASSCOOKIEID'];
    $check = mysql_query("SELECT * FROM `users` WHERE `username`='$user'") or die();
    if (mysql_result($check, 0, 'passwordcolnum') == $pass) {
    } else {
        //This is were the code goes for a user that is signed on
    }
} else { //what happens if they don't have the cookie
    header("Location: login.php");
}