这个特权准则有什么问题吗？

如何检查“检查”php代码或页面中的权限？

我使用爆炸和 in_array

用户登录并进入“检查”页面后，代码必须检查用户的权限是否具有“dataDisplay”权限。但“检查”页面中的代码不会执行此操作

我的“检查”页面代码有什么问题

这是我的数据库：

+--------------------+-------------------------------+
| username           |   user_privilege              |
|--------------------|-------------------------------|
| amal               |7gz,agt_courses,newbill        | 
|                    |                               |
+----------------------------------------------------+
|                    |                               |
| ahmed              |dataDisplay,previllige,newUsers|
+----------------------------------------------------+

第一页“登录”php：

<?php
ob_start();
session_start();
include '../connection/connect.php';

$username = $_POST['username'];
$password = $_POST['password'];


if($username && $password ){
    $finduser = mysqli_query($link,"SELECT * FROM LOGIN WHERE username = '".$username."' AND password = '".$password ."'") or die("error");
    if(mysqli_num_rows($finduser) !=0){
        while($row = mysqli_fetch_array($finduser)){
            $uname = $row['username'];
            $pass= $row['password '];
            $arr=explode(",",$row['user_privilege']);
        }
    }
        {
        $_SESSION['sessionname'] =$uname;
        $_SESSION['sessionpass'] =$password ;
        $_SESSION['sessionpre'] =explode(",",$row['user_previllige']);
        header ("location:../agtSite/agt2.php");
    }
} 
ob_end_flush();
?>

第二页“检查”php：

<?php
session_start();

$_SESSION['sessionpre']='';

$haspermission =in_array("dataDisplay",$_SESSION['sessionpre']);


if( $haspermission )
{
    header("location: agt2.php");
    }

else{header("location: ../display/display.php");}

?>

你忘了添加else块，如果正确缩进代码就可以避免这个问题。 确保仅在找到用户时才创建会话

if(mysqli_num_rows($finduser) !=0){
    while($row = mysqli_fetch_array($finduser)){
        $_SESSION['sessionname'] =$row['username'];
        $_SESSION['sessionpass'] = $row['password '];
        $_SESSION['sessionpre'] = explode(",",$row['user_previllige']);
        header ("location: ../check.php");
    }
}
else{
//could not find user
}

正如其他人在评论中指出的那样防止SQL注入 https://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php.