i am getting this error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/nanothre/public_html/guestbook/index.php on line 149

and this is the line of code it is referring to

PHP Code:

while($r = mysql_fetch_array($result, MYSQL_NUM)) 


It would be more helpfull if you would paste the lines around this line too.

sometimes an error is reported at line x but the fault is line y.

Listen to him/her ^^

The error means exactly what it says, $result is not a MySQL result resource meaning that there was probably an error with your query. As the messages above stated, try posting some more code to slim down your problem. When your executing your query its a good idea to stick some error handling in there as well, for example:

PHP Code:

$result = mysql_query("SELECT * FROM mytable") or die("MySQL Error: " . mysql_error()); 


Will display the mysql error, if there is one, and halt the script.

thanks for explaining what the error was, here is teh quesry code

PHP Code:

$query = "SELECT name, email, message, date, FROM guestbook order by id DESC LIMIT 10";
$result = mysql_query($query); 


thanks for the help guys, i re-built my query and it works now here is my new query

PHP Code:

$query = "SELECT `guestbook`.`name`, `guestbook`.`email`, `guestbook`.`message`, `guestbook`.`date`
FROM guestbook
ORDER BY `guestbook`.`id` DESC LIMIT 10";
$result = mysql_query($query); 


In case you wanted to know...

The problem was probably just because you have a comma before the FROM command which dosent need to be there.