this is a simple program it is not working...please do tell me whether my code is correct

function login()
 print "<script>";
print " window.location='http://www.webs.com/hi.html'";
 print "</script>";  
print "<input type='button' name='butt' value='Apply Online' onClick=login() style='height :2em; width :7em; font-weight:bold; font-size:0.7em' >";

you can't call a php function with an onClick event.. onClick is a javascript/DHTML item.

You also don't even need any of the php opening and closing stuff.
Why cause php to print javascript and html items if you already have form tags outside of the scope of the php?

function login(){
<input type='button' name='butt' value='Apply Online' onClick='login();' style='height :2em; width :7em; font-weight:bold; font-size:0.7em' >

Should do the trick.
Although, you really don't even need any of that stuff.
You are just wanting people to go to another page to fill out an application.
This is easily done in html

<a href="http://www.webs.com/hi.html"><img src="button.gif" /></a>


no if he is a registered user i want to redirect him to page1.php page or else page2.php...so i taught something like this will help me

ok, you still can't call a php function on a javascript event, which is what your code is trying to do.

What you need to do is use your php to print out the html code but make the onclick event itself the value.

Something like this.

if ($user == $registered){
$onclick = "window.location='http://www.webs.com/page1.html'";
else {
$onclick = "window.location='http://www.webs.com/page2.html'";
<input type='button' name='butt' value='Apply Online' onClick='
<?php echo $onclick; ?>
' style='height :2em; width :7em; font-weight:bold; font-size:0.7em' >

Obviosly you will replace user and registered with whatever you use to check users registration...