Question

Creating a login system for something and am getting:

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in …/func/user.func.php on line 21

Here is my code:

function user_register($email, $name, $password) {

}

function user_exists($email) {
  $email = mysql_real_escape_string($email);
  $query = mysql_query("SELECT COUNT('user_id') FROM 'users' WHERE 'email' ='$email'");

  //this is line 21:
  return (mysql_result($query, 0) == 1) ? true : false;
}

Answer 1

Your SQL is full of syntax errors, single quotes are used to quote string literals, backticks (or double quotes in standard SQL) are used for identifiers. Try this:

$query = mysql_query("SELECT COUNT(user_id) FROM users WHERE email = '$email'");

You don’t need to quote any of those identifiers so don’t bother.

From the fine manual:

For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error.

You probably want to add some error checking after you’ve fixed your SQL syntax errors.