Question

In the Java Servlet, how to include original paramters after response ?

Servlet
protected void service(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        String cmd = request.getParameter("cmd");
        System.out.println("service , cmd="+cmd);
        request.setAttribute("name", "John"+System.currentTimeMillis());
        RequestDispatcher rd = request.getRequestDispatcher("process.jsp");
        rd.include(request, response);
    }

JSP
main ${name}<br>
cmd ${cmd}

1. If I want to include all paramters, like “cmd”, to a new jsp page, how to do it ?
2. based on No.1, if I want to add NEW attributes, like “name” to a new jsp page, how to do it ?
3. In the above codes, use include or forward, the results are same. why ?

Thanks

Answer 1

If I want to include all paramters, like "cmd", to a new jsp page, how to do it ?

All request parameters are in EL available by the ${param} map.

${param.cmd}

You don’t need to prepare anything in the servlet.

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based on No.1, if I want to add NEW attributes, like "name" to a new jsp page, how to do it ?

You already did it by request.setAttribute("name", name) and ${name}.

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In the above codes, use include or forward, the results are same. why ?

Not exactly. If you use include(), the delegatee would not be able to control the response headers. You should be using forward() in this case. See also the javadoc. You should use include() only if you want to append something before and after instead of fully delegating the request/response.

See also:

• Our Servlets wiki page
• How do I execute multiple servlets in sequence?