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.
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