> -----Original Message----- > From: Jay Blanchard [mailto:jay.blanchard@xxxxxxxxxxxxxxxxxxxxx] > Sent: Thursday, May 12, 2005 9:37 PM > > 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; May? You _have_ to. This behavior was introduced in PHP4 as the only non-PHP3 compatible OOP behavoir as far as I know. Nice spotted for a beginner Rochelle :-) I mean the "I´m not allowed to use a function call while declaring with var". Explanation: PHP3 allowed one to use a function call while declaring a variable: var $welcome_text = "Welcome " . get_username_from_db($loginname) . ". Today is " . date("l") . ", have a nice day"; As of PHP4 one _can´t_ use a function call when declaring a var, but _has_ to do as in Your example Jay: var $welcome_text; $welcome_text = "Welcome " . get_username_from_db($loginname) . ". Today is " . date("l") . ", have a nice day"; Pay attention to this though: You _not_ suppose to use var to declare a variable unless it´s inside a class. I never really tested that before, but outside a class this won´t work: <? var $test = "hello"; echo $test; // returns nothing ?> -- Med venlig hilsen / best regards ComX Networks A/S Kim Madsen Systemudvikler/Systemdeveloper -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
