Understanding of url-patterns in web.xml

Question

I was really confused by this parts of web.xml:

<filter>
    <filter-name>filter</filter-name>
    <filter-class>com.labwork.filter.Filter</filter-class>
</filter>

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

This is the real problem. For example, I have several Servlets and static html pages. All request to them pass through filter. May be, i want to forward to servlet/html page from filter. How can i do this, if all request will be forward to filter? Or, may be, i don't understand principles.

All the request will be forwarded to the filter, then to your servlet/html if they passed the condition you provided on your filter. — lxcky
– lxcky, Commented May 10, 2017 at 0:11
@lxcky it provided by chain.doFilter? and how can i explicitly forward to servlet / html ? — Evgeniy
– Evgeniy, Commented May 10, 2017 at 0:23
Do you mean you want some request to go directly to their servlet/html pages and not go to the filter? If so, you should limit your url-pattern. /* means every request. — lxcky
– lxcky, Commented May 10, 2017 at 0:25
@lxcky No, I want all my request go throw filter. But i wan't specify in filter explicitly which servlet / html i want to forward. — Evgeniy
– Evgeniy, Commented May 10, 2017 at 0:29
Alright, let me compose an answer and see if it works out for you. — lxcky
– lxcky, Commented May 10, 2017 at 0:31

lxcky · Accepted Answer · 2017-05-10 00:46:40Z

1

Based on our conversation in the comment section, you want to filter every request with an exception of a few pages.

Let's say you want to exempt the login.html from filtering, what you could do is get the request URI and check if that string contains login.html like this:

public void doFilter(ServletRequest request, ServletResponse response, FilterChain chain) throws IOException, ServletException {                
    String path = ((HttpServletRequest) request).getRequestURI();
    if (path.contains("login.html")) //login page                                                
        chain.doFilter(request, response); //proceed to the page
    } else {            
        //conditions here
    }
}

I must state though that this is not the standard practice. If you're doing this then you should review your design choices.

edited May 10, 2017 at 0:46

answered May 10, 2017 at 0:41

lxcky

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ayip · Accepted Answer · 2017-05-10 01:23:27Z

1

If you want to specify which servlet/html the request should be forwarded to, you can add <servlet-name> and <servlet-mapping> sections in your web.xml.

For example:

<servlet>
    <servlet-name>LoginServlet</servlet-name>
    <servlet-class>examples.Login</servlet-class>
</servlet>

<servlet-mapping>
    <servlet-name>LoginServlet</servlet-name>
    <url-pattern>/login</url-pattern>
</servlet-mapping>

OR

<servlet>
    <servlet-name>LoginServlet</servlet-name>
    <jsp-file>/login.html</jsp-file>
</servlet>

<servlet-mapping>
    <servlet-name>LoginServlet</servlet-name>
    <url-pattern>/login</url-pattern>
</servlet-mapping>

answered May 10, 2017 at 1:23

ayip

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