> -----Original Message-----
> From: Jay Blanchard [mailto:[EMAIL PROTECTED]
> 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
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