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How to send values from one jsp page to other but redirect to some other page?

I want to send values from jsp1.jsp to jsp2.jsp but redirect jsp1.jsp to jsp3.jsp . I used the following code in servlet for jsp1
response.sendRedirect("welcome1.jsp");
request.setAttribute("usern",user);
RequestDispatcher rd = request.getRequestDispatcher("afterlogin.jsp");
rd.forward(request,response);
but it keeps on giving this error "org.apache.jasper.JasperException: PWC6033: Error in Javac compilation for JSP".

you can use the possiblites of sessions in those case so you can access the parameter in jsp2 and redirect the jsp 1 to 3
https://www.javatpoint.com/servlet-http-session-login-and-logout-example
http://java.candidjava.com/tutorial/Servlet-Jsp-HttpSession-Login-logout-example.htm

You can use a HttpSession object as a user session, or use the ServletContext object for sharing global application information. Then use methods getAttribute (String attb) and setAttribute (String attb, String value) for sharing information within JSPs. The issue when you use a request is that the domain of the request is constraining you to use this information only when you receive this request.
You can also use JavaBeans to share information within JSPs
EDIT: Have you included your Java code inside a scriptlet? Using <% your code here %>

when you are using the sendRedirect function it will immediately redirect to that page it will not read next line code.
So according to your code your compiler is smart enough to check it so it is giving you an error.
please see below
1) SendRedirect ():
This method is declared in HttpServletResponse Interface.
Signature: void sendRedirect(String url)
1)In case of sendRedirect request is transfer to another resource to different domain or different server for further processing.
2)When you use SendRedirect container transfers the request to client or browser so URL given inside the sendRedirect method is visible as a new request to the client.
3)In case of SendRedirect call old request and response object is lost because it’s treated as new request by the browser.
4)SendRedirect is slower because one extra round trip is required because completely new request is created and old request object is lost.Two browser request required.
5)But in sendRedirect if we want to use we have to store the data in session or pass along with the URL.
2) Forward():
This method is declared in RequestDispatcher Interface.
Signature: forward(ServletRequest request, ServletResponse response)
1)When we use forward method request is transfer to other resource within the same server for further processing.
2)In case of forward Web container handle all process internally and client or browser is not involved.
3)When forward is called on requestdispather object we pass request and response object so our old request object is present on new resource which is going to process our request
4)Using forward () method is faster then send redirect.
5)When we redirect using forward and we want to use same data in new resource we can use request. setAttribute() as we have request object available.
more in : https://www.javatpoint.com/q/3577/difference-between-requestdispatcher-and-sendredirect

Jersey: How to redirect/forward WITHOUT URL changing in address bar

I am using Jerysey's implementation of JAX-RS to write a REST API. My POJO that handles the get requests eventually forwards it to a JSP. Everything is working fine, the problem is that the forwarding causes the URL in the browser's address bar to change to the URL that the request was forwarded to. How do I do a redirect WITHOUT this URL changing in the address bar? Current, I have tried 4 different ways:
return Response.seeOther(uri).build(),
return Response.temporaryRedirect(uri),
//thrown exception:
throw new WebApplicationException(response),
return Response.status(303).location(uri).build();

It doesn't sound like a Jersey issue per se. Jersey is doing its part to receive the request, do some processing, and return the response you are expecting.
It sounds more like a servlet container issue. Why don't you want the url to change in the browser?
Restful services can (and should) be built with no concern about templates/JSPs/consumers. Take a look at a library like RestAssured, write some tests for your work, and you will see that it is acting as expected.
Instead of rendering out to a JSP, consider using a rest client to make straight http requests against your service.
If you want the url to remain unchanged, consider making the http call using an AJAX library (JQuery or other Javascript-based solution).
I hope that helps!

A RESTful resource is identified by a URL. So if you redirect to another resource with another URL the URL in the address bar should change. That's good because you can e.g. bookmark this URL or send it per eMail.
The question here is are you really redirecting to another resource or do you only want to return a different representation (HTML instead of e.g. JSON). If the latter you should not redirect. Let your resource-class directly return text/html by using Jerseys Viewables.

You could make the entire website inside an iFrame and load the new site into that frame. It will maintain the page URL and load your content.
http://www.w3schools.com/tags/tag_iframe.asp

Allow only "prerequisite" forwarding URL to access a servlet

Say for example I have a servlet named FooServlet, mapped to /foo.
There are two other servlets named BarServlet and CarServlet, mapepd to /bar and /car, respectively.
Now, if a user does a direct access to www.example.com/foo, they should be rejected. BUT if the request comes from a forward on /bar or /car, I will allow access to /foo.
Is this possible?

Yes, see this question: Java get referer URI?
You can check the referer header in the request to make sure they are coming from one of your other servlets.

You can use the request object to decide.
For ex :
Use request.getContextPath() to fetch the "/foo", "/bar" or "/car" and decide whether to allow the access or not.

Remove invalid request parameter from URL with servlet filter

I have a web application written in Java which uses Struts 1.0. Sometimes when a URL is fired, I can see that it one of the request parameters does not have any name and value like the following ...
http://www.aaa.com/test.do?a=1&b=2&=&d=4&e=5
As can be seen, there is a '&=' which is essentially a parameter with no name and value. I'd like to remove this part from the URL before sending the request to the server. How can I achieve this? Should I use a filter or is there an easier way?

How servlet filter will dispatch error message on request page?

I have written a servlet filter which is configured to be invoked for each url (/*). On the basis of some condition, if the condition is passed, I want to proceed normal execution by chain.doFilter(request,response), I also want to open same request URL with error message..
"say value entered in particular textbox is incorrect". Is this possible?
Do I have to use response.sendRedirect(request.getURL())? I hope I wont end up in infinite loop as I have configured filter on each URL. I am doing validation check on request parameter.

Just do the same as you'd do in a servlet: perform a forward.
request.getRequestDispatcher("/WEB-INF/some.jsp").forward(request, response);
A filter is by default not (re)invoked on a forward. Additional advantage is that the JSP reuses the same request and thus you can just set the validation error messages as a request attribute without the need for session or cookie based workarounds/hacks.
Unrelated to the concrete problem, this isn't entirely the right approach. Form-specific validation job should be performed in a servlet, not in a filter. If you'd like to keep your servlet(s) DRY, then look at the front controller pattern or just adopt a MVC framework which already offers a front controller servlet and decent validation out the box, such as JSF or Spring MVC.