Getting error when everything looks right.

[Smudly]

For some reason, the following code is giving this error:
Parse error: syntax error, unexpected '}' in /home/content/s/m/u/smudlys/html/ratestest.php on line 21

This is the line after the first if statement.

MySql version: 5.0.19

Here is my code:

<?php
/// Include Admin Page Here ///
include('inc/connect.php');

$user = $_SESSION['username'];
$getmember = mysql_query("SELECT * FROM users WHERE username='$user'");
while ($row = mysql_fetch_assoc($getmember))
{
$member = $row['member'];
}
$getlevel = mysql_query("SELECT * FROM userstats WHERE username='$user'");
while ($row = mysql_fetch_assoc($getlevel))
{
$level = $row['level'];
}

$addrate = .005;

if($member==0){
	$rate = ($level*$addrate+.50)
}
if($member==1){
	$rate = ($level*$addrate+1)
}

?>

[DexterMorgan]

Should it not be:

if($member==0){
    $rate = ($level*$addrate+.50);
}
if($member==1){
    $rate = ($level*$addrate+1);
} 

[Smudly]

That's it! Thanks alot

[bradgrafelman]

There are also lots of optimizations (and/or shorter code) that could be implemented... example:

<?php 
/// Include Admin Page Here /// 
include('inc/connect.php');

$user = mysql_real_escape_string($_SESSION['username']);
$sql = "SELECT u.member member, s.level level FROM users u LEFT JOIN userstats s ON u.username = s.username WHERE u.username = '$user' LIMIT 1";
$exec = mysql_query($sql);

if(($data = mysql_fetch_assoc($exec)) !== FALSE)
{
  extract($data);
}
else
{
  // display error message here - username not found in database
}

$addrate = .005; 

$rate = $level * $addrate + ($member ? 1 : .50);

Mainly note that a) there's no need to run separate queries when the data is related - use JOINs where applicable, and b) no sense in constructing a while() loop if you only expect at most one row to be returned.