MySQL not returning to my needs - php

Question

I am trying to create a function that is going to check if an user exists based on their email. After comparing the submitted email with the email entries in my database, I want to see if an user is legit or not. This is what I have coded so far:
function user_exists($email)
{
$email = mysql_real_escape_string($email);
$query = mysql_query("SELECT COUNT(`id`) FROM `imgit_users` WHERE `email`='$email'");
return (mysql_result($query, 0) <= 1) ? true : false;
}
And I used it like this:
if (user_exists($register_email) === true)
{
echo 'Fail';
$errors[] = 'The supplied email is already in use';
}
else
{
if (empty($errors))
{
$user_id = user_register($register_username, $register_email, $register_password);
$_SESSION['imgit_uid'] = $user_id;
echo 'Okay';
}
}
My script keeps returning Okay instead of Fail, even if I type in the same email everytime. Ideas?

Answer 1

return (mysql_result($query, 0) >= 1) ? true : false;
Swap the <= to >=

Answer 2

return (mysql_result($query, 0) <= 1) ? true : false;
So you want to return true if the result is less than or equal to 1. So if it's 1 or 0? Perhaps there is your error, to check if it is less than 1 and swap the true / false. Or check if it is equal to 1.

Answer 3

return (mysql_result($query, 0) >= 1) ? true : false;

Answer 4

I would prefer not use COUNT() to check if an email is already used, instead I'd use something like
function user_exists($email)
{
$sql = sprintf(
'SELECT 1 FROM imgit_users WHERE email = \'%s\' LIMIT 0,1',
mysql_real_escape_string( $email )
);
$result = mysql_query( $sql );
// add error handling
return 1 === mysql_num_rows( $result );
}