[SOLVED] Need help with authorization

[_absurd]

Ok, here is what I have so far in my config file:

 

if (empty ($_COOKIE['lhid']))
{
$user['auth'] = false;
$user[2] = 0;
}
else
{
$user = mysql_fetch_row (mysql_query ('SELECT `username`, `userid`, `level` FROM `users` WHERE MD5(CONCAT(`username`, `userpass`)) = \''.mysql_real_escape_string ($_COOKIE['lhid']).'\''));
if ($user)
{
	mysql_query ('UPDATE `users` SET `lastsec` = UNIX_TIMESTAMP(), `lastacip` = \''.$_SERVER['REMOTE_ADDR'].'\' WHERE `userid` = '.$user[1].' LIMIT 1');
	$user['auth'] = true;
}
else
{
	setcookie ('lh', null, 1);
	$user['auth'] = false;
	$user[2] = 0;
}
}

 

And here is where it is being used and receiving and error (line 3 below, line 30 in the error message):

"undefined variable: user"

 

function displayLogin()
{
	switch ($user['auth'])
	{
		case true:
			echo '<table class="small">
				<tr>
					<td class="head">Logged In</td>
				</tr>
				<tr>
					<td>Welcome, <b>'.$user[0].'</b>.<br /><br />
						<a href="#">[Contribute]</a><br />
						<a href="#">[Manage Account]</a><br />
						<a href="logout.php">[Log Out]</a></td>
				</tr>
			</table>';
			break;
		case false:
			echo '<table class="small">
				<tr>
					<td class="head">Login:</td>
				</tr>
				<tr>
					<td>Username:<br />
					<form name="login" method="post" action="login.php">
					<input type="text" name="username"><br />
					Password:<br />
					<input type="password" name="password"><br />
					<input type="submit" name="submit" value="Login"><input type="reset" name="reset">
					</form></td>
			</tr>
			<tr>
				<td><font size="1">Trouble? <a href="loginhelp.php">Login help.</a><br />
					No account? <a href="register.php">Register</a>.</font></td>
			</tr>
			</table>';
	}
}

 

But I defined it in the first bit of code, which is directly above the displayLogin function in the document. Can anyone help?

[MadTechie]

$user['auth'] wasn't passed to the function

 

try

 

function displayLogin($user)

 

if that makes sense

[_absurd]

Warning: Missing argument 1 for displayLogin()

 

Notice: Undefined variable: user

 

[ToonMariner]

or decalre $user as global within the function

 

function displayLogin()
{
global $user;
....


}

[_absurd]

Ah, that worked ToonMariner. Thanks 

[trq]

Globals should be avoided where possible IMO. You need to pass $user in to the function where you call it as well, just declaring the function excepts an argument is not enough.

 

Maybe take a look at the man on functions. this is pretty straight forward stuff.