Check number

Paragon

What I want to do is set up a user level check. I know this might be some what easy and there might be a tutorial. Please if anyone has the time help me out teach me how to do it. Also you could point me in the direction of the tutorial if there is one. Thanks alot.

OhLordy

Add an extra column to your user table (you could call it level, and it could be of type int(2)) then you need to put a level number in here when the user regiseters (this will be the base user level). All you have to do then is pull this level out when you log a user in and you have the users level. You can use this variable in a number of different ways.
I would use a number rather than a text code for the level as it is then easier to do more sweeping statements like

<?
if($USER->level > 3)
{
  header('Location: secret_page.php');
  exit();
}
else
{
  header('Location: normal_page.php');
  exit();
}
?>

What are you wanting to use the levels for?
Cheers
Rob

Paragon

posting comments and going to certain pages. I also was going to do the same thing for edit links. So a user could edit there own posts. I assume its done almost the same way?

Like if paragon == $poster or somthing?

Paragon

This is what I have so far its not working right though.
Does it look right to you?

<? 

$check = mysql_db_query($userdatabase, "SELECT DISTINCT $uid FROM users WHERE level='$level"); 


if ($check["level"] != 10)
{
	print "<font face=\"Arial, Helvetica, sans-serif\" size=\"5\" color=\"#FF0000\">";
	print "<b>Illegal Access</b>";
	print "</font><br>";
	print "<font face=\"Verdana, Arial, Helvetica, sans-serif\" size=\"2\" color=\"#000000\">";
	print "<b>You do not have permission to view this page.</b></font>";

	exit;
}
?>

tekky

No, its wrong

you cannot use WHERE on a column you never SELECTED

Weedpacket

Originally posted by tekky
you cannot use WHERE on a column you never SELECTED

You can't? Funny, it's never stopped me before...

Check the quotes in your query.

tekky

Originally posted by Weedpacket
You can't? Funny, it's never stopped me before...

Check the quotes in your query.

hrmmm didnt work when I tried... bah I'll refrain from SQL advice on little to no sleep

Paragon

I'm getting this error when I try it.

Unknown column 'shanksta' in 'field list'

$check = "SELECT DISTINCT $uid FROM users WHERE level='$level'";  


if ($check["level"] != 10) 
{ 
    mysql_query($check) or die(mysql_error());
    print "<font face=\"Arial, Helvetica, sans-serif\" size=\"5\" color=\"#FF0000\">"; 
    print "<b>Illegal Access</b>"; 
    print "</font><br>"; 
      print "<font face=\"Verdana, Arial, Helvetica, sans-serif\" size=\"2\" color=\"#000000\">"; 
    print "<b>You do not have permission to view this page.</b></font>"; 

    exit; 
}

drawmack

SELECT DISTINCT uid

not

SELECT DISTINCT $uid

Paragon

It will pick the right user then?

Unknown column 'uid' in 'field list'

OhLordy

"SELECT DISTINCT * FROM users WHERE level='$level' && uid='$uid'"

HTH

Weedpacket

You want to know if the user with uid $uid has a level of $level? All very simple SQL.

$query = "SELECT COUNT(*) FROM users WHERE uid=$uid AND level=$level"; // Just count them.
$result = mysql_query($query);
$count=mysql_result($result, 0, 0);
if($count>0)
{... // Yep, it's there
}

You want to know what level they have?

$query = "SELECT level FROM users WHERE uid=$uid";
$result = mysql_query($query);
if(mysql_num_rows($result)==0)
{ ... // That uid isn't even in the table.
}
else
{
    $level=mysql_result($result, 0, 0);
    if($level==10)
        {... // Level Ten User!
        }
}

I am of course assuming there is only one user with a uid of $uid. If there isn't, then it's not a very u id.

SELECT * is a last-ditch construct you use when you haven't the foggiest idea what there is and so you have to resort to asking for everything.

Paragon

Thanks alot man ill remember that. Ill try to use that code more so I learn what it does.

Paragon

I dont know if im doing somthing incorrectly, but its not working.


Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/shanksta/public_html/systemtest/admin/add.php on line 11

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/shanksta/public_html/systemtest/admin/add.php on line 18

Paragon

😃