PHP boolean error - php

Question

Im a newbie to php trying understand why i get boolean error in my specific case with a code that works for others. I have a function that returns a $user_id which is to be used in a session later on. Here is he function:
function login($username, $password) {
$user_id = user_id_from_username($username);
$username = sanitize($username);
$password = sha1($password);
return (mysql_result(mysql_query("SELECT COUNT (user_id) FROM users WHERE username = '$username' AND password = '$password'"), 0) == 1) ? $user_id : false ;
}
This always results in a fail, with this error: Warning: mysql_result() expects parameter 1 to be resource, boolean given in. with a line number that points to this line.
return (mysql_result(mysql_query("SELECT COUNT (`user_id`) FROM users WHERE username = '$username' AND password = '$password'"), 0) == 1) ? $user_id : false ;
Why does this fail ? Isn't this enough to check weather the query was successful and test result. DB connection and sql query seemed to be correct, doesn't matter whether i use ' or omit. After several hours of research i managed to remove the error with an if statement. Like so :
function login($username, $password) {
$user_id = user_id_from_username( $username );
$username = sanitize( $username );
$password = sha1( $password );
$result = mysql_query("SELECT COUNT (user_id) FROM users WHERE username = '$username'AND password = '$password'");
if ( $result == 1 ) {
return $user_id;
} else if ( $result == 0 ) {
return false;
}
}
I would like to know why the second function works and not the first. Do you have to use an if statement to check the query ? If there is a better way to write this function please suggest.:)

Answer 1

Probably you have error in query so mysql_query return false
Replace:
return (mysql_result(mysql_query("SELECT COUNT (user_id) FROM users WHERE username = '$username' AND password = '$password'"), 0) == 1) ? $user_id : false ;
with:
$result = mysql_query("SELECT COUNT(user_id) as count FROM users WHERE username = '$username' AND password = '$password'") or die(mysql_error());
return (mysql_result($result, 0, 'count') == 1) ? $user_id : false ;
and you will see mysql error.