Question

我有这段代码但是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）

if(isset($_COOKIE['USERCOOKIEID'])) { 
$user = $_COOKIE['USERCOOKIEID']; 
$pass = $_COOKIE['PASSCOOKIEID'];
$check = mysql_query("SELECT * FROM users WHERE username = '$user'")or die();
while($info = mysql_fetch_array($check)) {
    if ($pass != $info['password']) {           
    }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");

}

由于

Answer 1

应该看起来像

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，代替mysql_fetch_array，为什么不使用mysql_result，因为肯定只会有一条记录

这段代码出了什么问题[PHP Cookies]

1 个答案: