Check if value is in mysql's table's column - php

Question

How can I check if value is in mysql's table's column?
Right now I'm creating a login system. After the user registers my php file posts the value and makes it into a variable. I was wondering what code i can use to compare my value to those values in the table and to check whether or not there is a value identical to the value entered.
I tried this code.
$query = "SELECT * FROM users WHERE username ='$username';";
$res = mysql_num_rows(mysql_query($query));
and then if ($res >= 0);
but then i get this error,
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/synameg1/public_html/giveaway/appfiles/register.php on line 23
so i was wondering if there is either a way to fix this error or do what i want to do in another method.
Thanks!

Answer 1

$sql = "SELECT * FROM users WHERE username = '" . mysql_real_escape_string($username) . "'";
$result = mysql_query($sql) or die("Error occurred in [$sql]: " . mysql_error());
$count = mysql_num_rows($result);

Answer 2

Dan: I used your approach with 'zero' result.
<?php
//////////////////////////////////////////////////////////////////////
global $current_user;
get_currentuserinfo();
$user_nam = $current_user->user_login;
$db_host = 'localhost';
$db_user = 'Private'; ////the user name
$db_pwd = 'Private'; /////the pasword
$database = 'Datapase name';/////////
$table = 'Table Name'; ///table name
$sql = "SELECT * FROM $table WHERE 'Submitted Login' = '" . mysql_real_escape_string($user_nam) . "'";
$result = mysql_query($sql) or die("Error occurred in [$sql]: " . mysql_error());
$count = mysql_num_rows($result);
echo $user_nam . "<br />";
echo "count is $count";
////////////////////////////////////////////////////////////////////
?>
Private fields have comments.
It works, I can see $user_nam but the count is zero.
Here is a datebase peek so you can see the values.
http://you-get-real.com/images/database-peek.png
Thanks