Display forwarded JSP with url-pattern "/*"

Question

To improve my java skills, I'm trying to build a simple j2ee framework (MVC).

I built it to handle every request in a FrontServlet. Here is the mapping that I used :

web.xml :
<servlet>
    <servlet-name>Front</servlet-name>
    <servlet-class>test.FrontServlet</servlet-class>
</servlet>
<servlet-mapping>
    <servlet-name>Front</servlet-name>
    <url-pattern>/*</url-pattern>
</servlet-mapping>

My problem is that when I forward the request from the FrontServlet to a JSP, obviously, the JSP request is handle by the FrontServlet and the view isn't rendered.

How can I resolve this problem by keeping the url-pattern "/*" ?
Is there a way to render a JSP in a Servlet without performance losses ?

Thanks in advance for your reply !

Solution 1 (@Bryan Kyle)

I'm trying to follow your advise. I created this filter :

public void doFilter(ServletRequest request,
   ServletResponse response, FilterChain chain) 
   throws IOException, ServletException 

   {
       HttpServletRequest req = (HttpServletRequest) request;

       if(!req.getRequestURL().toString().endsWith("jsp"))
       {
           // I changed the servlet url-pattern to "/front.controller"
           req.getRequestDispatcher("/front.controller").forward(req, response);
           /*chain.doFilter(req, resp);*/
       }
   }

<filter>
    <filter-name>Filter</filter-name>
    <filter-class>test.Filter</filter-class>
</filter>

<filter-mapping>
    <filter-name>Filter</filter-name>
    <url-pattern>/*</url-pattern>
</filter-mapping>

Is it right?

Thanks !

Community · Accepted Answer · 2017-05-23 12:08:16Z

1

A Filter is an inappropriate solution for a front controller approach.

You want to refine the url-pattern of your servlet so that it matches e.g. /pages/* or *.do. You don't want your front controller to kick in on irrelevant requests like CSS/JS/images/etc. To take /pages/* as an example, assuming that you've a JSP in /WEB-INF/foo.jsp, then the following in a servlet

protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    request.getRequestDispatcher("/WEB-INF" + request.getPathInfo() + ".jsp").forward(request, response);
}

should display the JSP in question on http://localhost:8080/contextname/pages/foo.

1 Comment

Pierre Over a year ago

Yes, it's probably the easiest way to resolve the problem, thank you !

Bryan Kyle · Accepted Answer · 2011-02-23 19:38:14Z

1

I think the problem here might be that you're using a Servlet instead of a ServletFilter.

A ServletFilter, as the name suggests filters requests by providing pre- and post-processing on the request. You'd probably want to use a Filter if you needed to do something like the following:

Provide security checks across an entire application
Set request properties that are picked up by a servlet or jsp
Compress a response
Log timing information
Etc.

Have a look at the documentation about Servlet Filters.

answered Feb 23, 2011 at 19:38

Bryan Kyle

13.8k4 gold badges42 silver badges45 bronze badges

1 Comment

Pierre Over a year ago

It could be a solution. I will try and I will edit my question

Collectives™ on Stack Overflow

Display forwarded JSP with url-pattern "/*"

2 Answers 2

See also:

1 Comment

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

See also:

1 Comment

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related