[snip] Thank you for your response Jay, but that is not working. My program will not run at all with the following: var $exportFile = "Export." . date("mdy") . ".txt"; I seem to be able to use the date function is I am not starting the declaration with "var", but then my program is not working correctly. [/snip] You may have to assemble it beforehand sort of ... $exportFileName = "Export." . date("mdy") . ".txt"; var $exportFile = $exportFileName; -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php