[SOLVED] little help

[Minase]

hy there lets say that i have a code that is "problematic"

here it is

$q = "SELECT * FROM `users` WHERE `Username` = 'Minase' ";
$test =  mysql_query ($q);
$test2 = mysql_num_rows($test);

this is not the real code,i did split the code maybe it was something weird about it,but it isnt.

the problem is that it return

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in LOCATION

 

my real code

function count ( $query )
	{
		return mysql_num_rows ( mysql_query ( $query ) );
	}

 

any idea why it return that error?  thank you

[MasterACE14]

I don't think you can call your function 'count' as that is already taken.

[Minase]

it does not matter the name,it was just informative,the problem is that even the simple querys are not working 

[Philip]

function count ( $query ) {
        $q = mysql_query ( $query ) or die(mysql_error());
return mysql_num_rows ( $q );
}

 

I would imagine it's because you cant connect to the db. But lets see what it dies.

 

Oh and as mentioned, be careful using the same function names.

[Minase]

oh baka me thank you solved the problem