[RESOLVED] mysql_result(): supplied argument is not a valid MySQL result resource

[tymilligan]

I am trying to check if a code is already present in mySQL server and when I run it, it gives me this error: "mysql_result(): supplied argument is not a valid MySQL result resource". Would I need to use a mysql_query();?

Here is the code

<?php
include 'db.inc.php';
function is_chlink($url) {
	return (preg_match("/chlink\.co\.cc/i", $url)) ? true : false;
}
function gen_code() {
	$charset = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789';
	return substr (str_shuffle($charset), 0, 6);
}
function code_exists($code) {
	$code = mysql_real_escape_string($code);
	$code_exists = mysql_query("SELECT COUNT(url_id)FROM url WHERE code='$code' LIMIT 1");
	return (mysql_result($code_exists, 0) == 1) ? true : false;
}	
?>

[johanafm]

tymilligan;10997300 wrote:

error: "mysql_result(): supplied argument is not a valid MySQL result resource".

Search the forum and use one of hundreds of answers to it. They are all the same.

[bradgrafelman]

There's also a link in my signature.

The gist of it is that you need to be checking if [man]mysql_query/man indicated an error has occurred and, if so, either output or log the MySQL error message (e.g. by using the [man]mysql_error/man function).

And then there's the issue of the entire [man]mysql[/man] extension being deprecated in favor of the newer/improved [man]MySQLi[/man] extension...