i'm going nuts

hi,

i'm new to php mysql and already have a problem with a simple routine

$result=mysql_db_query(jcorp,"select count(*) as bar from users");
$count = mysql_result($result,0,"bar"); //<<<<< this is line 3 in the
script
echo "Active Members $count";


this routine is working fine local coudn't be better

but when i put it on my site i get the following error

Warning: mysql_result(): supplied argument is not a valid MySQL result
resource in /home/virtual/site755/fst/var/www/html/active.php on line 3


what is wrong here???? i searched the web for different routines all work
localy but non work on the site they all give much the simelar error

please help me,
tnx

"The Lone Wolf" <jcorp.webmaster@pandora.be> wrote
in message news:<KSUZa.62130$F92.7199@afrodite.telenet-ops.be>...
>
> i'm new to php mysql and already have a problem with a simple routine
>
> $result=mysql_db_query(jcorp,"select count(*) as bar from users");
> $count = mysql_result($result,0,"bar"); // this is line 3 in the script
> echo "Active Members $count";
>
> this routine is working fine local coudn't be better
>
> but when i put it on my site i get the following error
>
> Warning: mysql_result(): supplied argument is not a valid MySQL result
> resource in /home/virtual/site755/fst/var/www/html/active.php on line 3

Is jcorp on Line 2 really a constant?

Cheers,
NC