--- Burak Delice <[EMAIL PROTECTED]> wrote:
http://localhost/menu.php?status=0 php return
[snip]
if ($status==0) echo '"trying";
if (!$_GET['status']) { echo '"trying"; }
Firstly, !$_GET['status'] means "if status is not equal to or greater than zero" -- which of course will not work.
Also, this will not work because 0 is considered false in PHP. So if status is 0, then it won't echo anything (nevermind the syntax error).
Try
$status = isset($_GET['status']) ? $_GET['status'] : NULL;
echo $status;
-- Burhan Khalid phplist[at]meidomus[dot]com http://www.meidomus.com ----------------------- "Documentation is like sex: when it is good, it is very, very good; and when it is bad, it is better than nothing."
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