Try:
if (isset($_GET['status']) && $_GET['status'] ==0) {
echo "Trying";
}
The warning is triggered by the fact that you use a variable that was not
previously defined. Check it before use with isset.
If in php.ini the value error_reporting = E_ALL, it will output warnings.
If you want the warnings to dissapear, set in php.ini:
error_reporting = E_ALL & ~E_NOTICE
Lucian
"Burak Delice" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> hi,
>
> when I enter url like that http://localhost/menu.php?status=0 php return
an
> error below :
> "Warning: Undefined variable: status in
> D:\calismalar\cengiz_ozdelice\web\menu.php on line 41"
> my code on 41.line is :
> if ($status==0) echo '"trying";
>
> what is problem?
>
> thank you.
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