can some on take a look at this code?

[gat]

I get the error

------------------
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\program files\apache group\apache\htdocs\ky\walls.php on line 197
-----------------

and the code look like

-----------------------------
<?php

$conn= @mysql_connect("localhost","*****","*******")
or die("Err:Conn");

$rs= @mysql_select_db("kriminalyouth", $conn)
or die("Err:Db");

$query = mysql_query ( "SELECT COUNT (*)AS number FROM walls_berks") ;

list($count) = mysql_fetch_array($query);

echo "there are " . $count . " images in the table";

?>
---------------------------

many many thanks in advance

[liquedus]

i would put something like

if(!$query) print "MySQL error " . mysql_error();

after mysql_query()

if there is a problem with your query this would show it. which is line 197?

[gat]

197 = list($count) = mysql_fetch_array($query);

hope this helps.