Not a valid result resource?

[gergy008]

	$run=mysql_query("SELECT `verifid` FROM `users` WHERE `login`=$user", $link);
	if(mysql_num_rows($run)==1){

 

These lines seem to be giving me the 'Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource' error. Looks fine to me, But probably my fault.

 

Thanks in advance.

 

Here is the whole function if you need it:

function getverid($user){
	global $link;
	$run=mysql_query("SELECT `verifid` FROM `users` WHERE `login`=$user", $link);
	if(mysql_num_rows($run)==1){
		$item=mysql_fetch_array($run);
		echo($item[0]);
		return($item[0]);
	}
}

 

The fuction is called with getverid($_POST["login"]) which I know is set... So what's going on?

[kenrbnsn]

You have a syntax error in your query. If you replace

<?php
$run=mysql_query("SELECT `verifid` FROM `users` WHERE `login`=$user", $link);
?>

with

<?php
$q = "SELECT verifid FROM `users` WHERE `login`=$user", $link";
$run = mysql_query($q) or die("Problem with the query: $q<br>" . mysql_error());
?>

It would show the error.

 

The error is WHERE `login`=$user", $link -- what's the , $link? (nevermind)

 

Ken

[-Karl-]

$link is probably the problem, show more of the code or try to debug the query.

 

 

 

EDIT:

 

Okay, I can see where $link is now. Like kenrbnsn said, debug the query to show the error.