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Prevent form resubmit after pressing back button

I am in bit of a delicate situation here. In my organization we design stock management systems and it is a web application based on JSP pages and servlets which handles them.
I have been asked to fix a specific problem. We have a JSP page with an HTML form table where there are stock details. When user enters the details manually and submit the form, stock details updated in the database and it works fine.
Problem is this : When the user press the browser's back button, user can come to the previous page where he submitted the details. And when the user submit this, data is saved once more to the database.I need to prevent this behaviour.(Something likeclear and reload the page.)
Things I did so far : clear the browser cache.Code works fine but not the expected result.
Unfortunately I cannot share the code due to company regulations. What I need is a help to prevent this behaviour or a workaround.
Thanks in advance..

You can use a javascript function with the help of a hidden attribute to reload the web page. When the user press the back button,based on the value of the hidden attribute, page will be reloaded without loading the cached page.
Your approach of clearing cache is correct. Coupled with that, you can use this approach.
<input type="hidden" id="refreshed" value="no">
<script type="text/javascript">
onload=function(){
var e=document.getElementById("refreshed");
if(e.value=="no")e.value="yes";
else{e.value="no";location.reload();}
}
</script>
One drawback of this approach is if your clients' browsers have disabled JS, this will not work.Otherwise it should work.

When the user press the browser's back button, user can come to the
previous page where he submitted the details. And when the user submit
this, data is saved once more to the database.
According to how you described it, that is based on a doGet request. Which means every time you visit that URL, it will send the request with whatever parameters were added.
As someone already mentioned, if you switch the form to a post method and switch the Servlet to a doPost, you won't have this issue anymore.
Alternatively you can circumvent this with a javascript solution. Here are some options:
You can check if the user clicked the back button, disable form if true.
Another way is by storing a cookie which you check on page load, if it exists you can disable the form.

You can use this code also
$(document).ready(function() {
function disableBack() { window.history.forward() }
window.onload = disableBack();
window.onpageshow = function(evt) { if (evt.persisted) disableBack() }
});

You must use a Post-Redirect-Get pattern: https://en.m.wikipedia.org/wiki/Post/Redirect/Get.
Actually, every use of standard HTML forms with method="post" should be implemented with that pattern. It doesn't have any use for AJAX-posted forms, which actually could be another solution but will require more work and probably some architectural changes.

I had this same problem while building a django web app, and my solution was to not allow caching of the html that contains the form. In your request handler, do not allow the browser to cache the page. This will force the browser to get the page fresh from the document.
Which, in this case, you can just verify in your request handler if the requested form has already been submitted.
My code for reference:
from django.views.decorators.cache import never_cache
#never_cache
def GetForm(request, pk):
# Logic #
if (IsFormCompleted(pk)):
# Handle request #

Here is a solution.
give a random id in a hidden field on the form. Then on the server side, if the user resubmit, check if the random id already on the database. If so, redirect user.

How to grab a whole body unless header and footer of another site page

I'm developing with Liferay portal.
And now I'm facing a little problem:
I'm making site for some Company that has subsidiaries.
Then, I must cut out some parts(precisely header and footer)
of other site(sub. site) and put the body of page without'em in iframe of main site.
I was "googling", looking for something about Grabbers.
but I've found just about how to grab with PHP or Perl.
and here
It doesn't seem to be exact what I need.

You can try the WebProxy portlet for this.
As you'll have to modify the external content's body, you can't simply show it in an iframe, so this portlet might be what you need. It doesn't work with an iframe internally and you can replace some content on-the-fly.

Liferay Page redirection

I have developed a theme in liferay 6.1. I have a page named "localhost:8080/home" but now i want that on clicking this link of the page, it should be redirected to localhost:8080
Any suggestions are welcomed.
Thanks in Advance.

I think you are confused a little bit, so just some things you should know:
You can't (normally and without hacks) have a page named "localhost:8080". Every Page (or 'Layout' in Liferay) has a short name, that takes it's part of the url. This is often called "friendly url" but it's often confused with the "friendly url feature", which is a way to shorten your url request data.
So you're always going to have urls like 'localhost:8080/something'. The same holds for the 'home' page
You can partially shorten the Url by using 'virtual host'. It removes the part of the url before your page's name (like removing the web/guest or user/username ) suffix
You can use the 'friendly url' feature to shorten the part of the url that goes after the page's name, and contains request information like lifecycle state info or custom request parameters

How to change url in user browser without client redirect in servlet

I want to forward from one page to another but with the same I want url to be changed. Suppose user is here http://mywebsite/register and when he completes his registration process then I want this in his address bar http://mywebsite/home
Is it possible without using sendRedirect , I mean by the way server side forwarding only? or any other way around to this problem?

You could just let the HTML form submit to that URL directly.
<form action="http://mywebsite/home">
But this makes no sense. You'll also run into problems when redisplaying the same form with validation messages in case of validation failure. You'd need to redirect back to the original page if you intend to keep the original URL and you'd need to fiddle with storing messages in the session scope instead of the request scope because a redirect basically creates a brand new request. You'll without a redirect also run in "double submit" problem whenever the enduser presses F5 after submitting the form.
Just let the servlet redirect the successful POST request to the desired URL. That's the canonical approach. Even more, this is a recommend "design pattern": the POST-Redirect-GET pattern.

AFAIK there's no way around a redirect since the browser has to update the url at some point. And if you'd update the url after the forwarded to page has been loaded it would issue a refresh and the page would be loaded again (which might result in an endless loop).
Why don't you want to use a redirect in that case?

Is it possible to mask/ hide the lengthy URL and just to display the domain name alone?

Is it possible to mask/ hide the lengthy URL and just to display the domain name alone in address bar in the browsers like IE, Firefox, Chrome?
Please suggest.
Regards
Gourav

You shouldn't do that.
It's against the very basics of the technology and usability.
Every page should have its unique address, letting users bookmark it, send link to a friend, navigate your site after all!

You want to use AJAX for this. In your index file, include a javascript file which uses XMLHttpRequest (or you can use something like jQuery.load if you don't want to go so low level) to load your content. With jQuery, you can do something like this:
<script type="text/javascript">
$(document).ready(function(){
$('a').click(function(event){
event.preventDefault();
$('#content').load($(this).attr('href'));
return false;
});
});
</script>
However, do not do so lightly - this can break search engine optimization and many other things as some people may have javascript off etc.
And remember, this is a very simplified example - you'd have to take care of things like external URIs (CSRF protection in browsers means you can't XMLHttpRequest another domain). Maybe you could add a CSS class called link_internal and then add that in your jQuery selector etc.

domain = re.match(r'https?://(?:www\.)?([^/]+)', full_url).group(1)
This regex extracts the domain - without www. but any other subdomains if they exist.
It uses the python re module but it should be easy to adept it to another language.