> -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 i
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 t
hmmm, well... If I could see the code I could probably fix it for you,
the problem is everything changes depending on the context it's used
and how the class is setup. For example, it's poor practice to assign
values to vars when you declare them... It is recommended you assign
them inside the cla
On 5/12/05, Clinton, Rochelle A <[EMAIL PROTECTED]> wrote:
> Dear Much Needed Advisor,
>
> I am definitely a PHP novice and making some code changes to a PHP
> script.
>
> I need to change a declared output file to include the date as an
> extension.
>
> My file is declared at the beginning of t
[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 c
Will this work?
-Original Message-
From: Clinton, Rochelle A [mailto:[EMAIL PROTECTED]
Sent: Thursday, May 12, 2005 12:55 PM
To: php-general@lists.php.net
Subject: [PHP] beginner needs help!
Dear Much Needed Advisor,
I am definitely a PHP novice and making some code changes to a P
Clinton, Rochelle A wrote:
Dear Much Needed Advisor,
I am definitely a PHP novice and making some code changes to a PHP
script.
I need to change a declared output file to include the date as an
extension.
My file is declared at the beginning of the script as:
var $exportFile = "Export.txt";
I need
[snip]
My file is declared at the beginning of the script as:
var $exportFile = "Export.txt";
I need it to be, for instance, Export.051205.txt.
I have played around with the date function many ways:
Ex 1:
var $exportFile = "Export" . date("mdy");
[/snip]
You almost had it;
var $exportFile =
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