On Tuesday 08 June 2004 19:13, php-general wrote:
> $var3 is defined as
>
> $var3 = "$var1$var2";
>
> in the body of the if ($button1) statement !?
what is button1 supposed to do and what is button2 supposed to do?
--
Jason Wong -> Gremlins Associates -> www.gremlins.biz
Open Source Software Sy
On Tuesday 08 June 2004 18:50, php-general wrote:
> Only pressing the "Display concatenated strings"-button should
> output the concatenated string, but this does not work, and I do not
> understand why.
$var3 is not defined.
--
$var3 is defined as
$var3 = "$var1$var2";
in the body of
On Tuesday 08 June 2004 18:50, php-general wrote:
> Only pressing the "Display concatenated strings"-button should output the
> concatenated string, but this does not work, and I do not understand why.
$var3 is not defined.
--
Jason Wong -> Gremlins Associates -> www.gremlins.biz
Open Source S
- Original Message -
From: php-general
To: [EMAIL PROTECTED]
Sent: Tuesday, June 08, 2004 12:36 PM
Subject: Problems with variable handling !
Hi !
I have a Problem with hand over of variables;
My Page consists of three files which I have enclosed to
demonstrate the
phpinfo reports that register_globals=On
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