2013-10-17 271 views
-2

我通過implode向user_privilege列添加值,包括用戶的權限,即 (7gz,agt_courses,newbill,new_Trainee)=>他們每個人都可以打開(例如:newbill頁面)或不打開在用戶權限 user_privilege列在數據庫 「登錄」表中的數據庫 換句話說,我想檢查用戶登錄他的權限,如果它有說新手權限,他可以打開新手頁面,如果不是他不能打開此頁面。檢查用戶權限的權限

登錄表數據庫是這樣的:

**username  username2   user_privilege** 

amal    amal   7gz,agt_courses,newbill 

ahmed   ahmed   dataDisplay,previllige,newUsers 



$username = $_POST['username']; 
$username2 = $_POST['username2']; 

if($username && $username2) 
{ 
    $finduser = mysqli_query($link,"SELECT * FROM LOGIN 
    WHERE username='".$username."' AND username2='".$username2."' AND FIND_IN_SET('Customers', user_previllige)") or die("error"); 

    if(mysqli_num_rows($finduser) !=0) 
     { 

      while($row = mysqli_fetch_array($finduser)) 
       {$uname = $row['username']; 
       $uname2 = $row['username2'];} 
     } 

     if($username == $uname && $username2 == $uname2) 
      {$_SESSION['sessionname'] =$uname; 
      $_SESSION['sessionname2'] =$uname2; 
      header ("location:../customers/cutomer.php");} 
       else echo '<script>function 
{alert(you have not permission to open this page)}</script>'; 

    } 

任何機構幫助我?

+0

哪裏是什麼,你已經嘗試過的代碼? –

+1

我們如何能夠幫助您,而不會看到數據是什麼樣的? – rorypicko

回答

1
function check($sPermission, $sColumnValue) 
{ 
    $aPerm = explode(',', $sColumnValue); 

    return (!empty($aPerm) && in_array($sPermission, $aPerm)); 
} 

我希望你知道如何從數據庫中選擇數據,所以調用將是這樣的:

check('newbill', '7gz,agt_courses,newbill');