Re: giving variable value in URL problem

Try: if (isset($_GET['status']) && $_GET['status'] ==0) { echo "Trying"; } The warning is triggered by the fact that you use a variable that was not previously defined. Check it before use with isset. If in php.ini the value error_reporting = E_ALL, it will output warnings. If you want the warnings to dissapear, set in php.ini: error_reporting = E_ALL & ~E_NOTICE Lucian "Burak Delice" <bdelice@yahoo.com> wrote in message news:20031116013209.51850.qmail@pb1.pair.com... > hi, > > when I enter url like that http://localhost/menu.php?status=0 > php return an > error below : > "Warning: Undefined variable: status in > D:\calismalar\cengiz_ozdelice\web\menu.php on line 41" > my code on 41.line is : > if ($status==0) echo '"trying"; > > what is problem? > > thank you.