Login page according to the access rights

ghjk

In my web site I want to load some menus without login and there is a admin login. When admin logged in I want to some more menus. This is my code. But it is not working. It display all menus every time. Please help me.

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$page = $_GET['page'];
 
if ($page);

else 

$page = "index.php"; 
 
    if($login=="True"){

    //echo "inside admin login";

        $array = array('Home'=>'index.php','Stock'=>'stock.php','Order Now'=>'orderNow.php','Contact Us'=>'contectUs.php','Add'=>'addVehicle.php','Edit'=>'editVehicle.php','Logout'=>'logout.php');
 
    foreach ($array as $k=>$v){

        if ($page == $v)

            $background = "background='images/btn_inactive.gif' ";

        else

            $background = "background='images/btn_active.gif'";
 
        echo "<td width=200 height =35 align=center cellpadding=5 $background ><a href='$v?page=$v' class=toppanel >&nbsp;&nbsp;&nbsp;$k </a></td>";
 
        }
 
    }

Dormilich

how do you determine whether an admin or an user is logged in?

TheServant

It looks like your line 8 that is failing if it shows all the items.
Instead of checking if login is true, which you should also check if the type is the same like this:

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if ($login===TRUE) {

You could also check if a certain variable isset() or if a $_SESSION value (which I presume is how you know they're logged in) is set?

ghjk

In my login page I checked whether user logged in or he is a guest.Actually my site no need to have any user login. I want to do a login part only the admin to enter data.
This is my login page and I'll get the value "true" from here.

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if(isset($_POST['login'])){

    $username=$_POST['username']; 

    $password=$_POST['password'];

    if((!username)||(!$password)){

        $login_error = '<p class= "errormsg">Please enter all fields</p>';

    }

        $password_encrpt = md5(substr($password, 1, -1));

        $sql = "SELECT * FROM user WHERE  Username='$username' AND Password ='$password_encrpt'"; 

        $result=mysql_query($sql);

        $row = mysql_fetch_array($result);

        $count=mysql_num_rows($result);

        if($count==1){

            echo $_SESSION['Username_Ses'] =$row['Username'];

            $_SESSION['login'] = "True";
 
        }

        else {

            $ADDERROR = '<p class="errMsg" align="Center">Invalid Username or Password</p>';

        }
 
}

TheServant

In your original script you do not use $_SESSION['login'], you use $login, which does not exist.

Login page according to the access rights

re: Login page according to the access rights

re: Login page according to the access rights

re: Login page according to the access rights

re: Login page according to the access rights

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