Re: value error in PHP form

Maybe this might work .... ---------- <html> <head> <title></title> </head> <body> <?php # check if the form has been submitted and action = login if(isset($_POST['action'])&&$_POST['action']=="Login") print $_POST['action']; print $_POST['Name']; # more statements ... ?> <form name="frmname" action="<?php print $_SERVER['PHP_SELF']; ?>" method="post"> <input type="text" name="Name" value="Rinku" /> <input type="submit" name="action" value="Login" /> </form> </body> </html> ---------- ----- Original Message ----- From: "Rinku" <rinku_pgsql@yahoo.com> To: "Marvin Hechanova" <mhechanova@gmail.com>; <php-db@lists.php.net> Sent: Sunday, June 20, 2004 2:18 PM Subject: Re: [PHP-DB] value error in PHP form > Actually I want to use the function like > if($action="Login") > { > Statements; > } > But here I am not getting any value in $action even I click on it. > When I was using Linux at that time I had not this kind of problem. > > Rinku > Marvin Hechanova <mhechanova@gmail.com> wrote: > You have to assign values to your name and action > e.g. > > [input] " value=Rinku> > [input] " type=submit value="Submit Query" > value="Login"> > > but why would you want to print the results? > > > On Sun, 20 Jun 2004 04:22:59 -0700 (PDT), Rinku wrote: > > > > Dear All, > > > > I have installed PHP on WinXp. I am using MySql as Backend on Apache server. > > Now the problem is...... > > I am writing this code : > > > > > > > > > > > > > > > > > > [input] > > [input] > > > print $action; > > print $Name; > > ?> > > > > > > > > Here I should get output as LoginRinku > > But I am getting nothing. > > > > Can any of you guide me on this........? > > > > Regards, > > Rinku > > > > > > --------------------------------- > > Do you Yahoo!? > > New and Improved Yahoo! Mail - Send 10MB messages! > > > --------------------------------- > Do you Yahoo!? > Yahoo! Mail - 50x more storage than other providers!