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;
}
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.
There is an error in your query. If you want to figure out what the error is, output it with mysql_error()
Your query has syntax errors. You're using single quotes on field and table names. That changes from them being field/table names to being ordinary strings. So your query boils down to count(some string) from someotherstring where yetanotherstring.
If your query call had been constructed something like this:
$result = mysql_query(...) or die(mysql_error());
you'd have been informed of the syntax error. As it stands now, your code assumes the query succeeds, which is a very bad thing to do. There's precisely ONE way for a query to succeed, and far too many ways for it to fail.
Your code has no error handling, so if anything goes wrong, it continues on blissfully unaware trying to make soup out of stones. Take a look at the examples in the manual, they all check for errors in the query before they try to analyze the result.
