RE: [PHP] beginner needs help!

On Thu, May 12, 2005 12:36 pm, Jay Blanchard said: > [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 has no 'var' at all, anyway, *EXCEPT* inside a 'class' definition, and PHP5+ has better keywords (public/private/ etc) to use instead of 'var' Furthermore, where you do have var in PHP, you can ONLY provide it with a constant. It cannot be a derived value. Not a variable. Not an expression. A constant. Disclaimer: I don't use classes, so maybe I'm wrong... But I think I got this one right. -- Like Music? http://l-i-e.com/artists.htm