CODEWITHSUNDEEP
javaservletsweb.xmlurl-mapping
guogangj
guogangj

Reputation: 2425

How to make servlet map a special url to a static resource?

I know, by the default servlet, static resources will be returned automatically. However, my url is quite special and I don't want it looks like http://mysite/app/test.html. Maybe it is like that:

http://mysite/app/dosomething/7419698f

I want to map(or forward?) this request to a static html file, for example /WEB-INF/pages/dowork.html. For more, 7419698f is only a parameter, http://mysite/app/dosomething/2926739e will also be mapped to the same static file. I know a workaround like that:

<servlet>
    <servlet-name>test</servlet-name>
    <jsp-file>/pages/dowork.html</jsp-file>
</servlet>
<servlet-mapping>
    <servlet-name>test</servlet-name>
    <url-pattern>/dosomething/*</url-pattern>
</servlet-mapping>

It works only if the dowork.html is a legal jsp file. If I want to serve image resource like that, it would turn error.

Upvotes: 2

Views: 1246

Answers (2)

guogangj
guogangj

Reputation: 2425

Leos Literak's clue is right. My requirement cannot be done by simple xml config, I must do something more.

I need a filter. In web.xml:

<filter>
    <filter-name>summary-fw-filter</filter-name>
    <filter-class>com.mycompany.mywebapp.filter.SummaryForwardFilter</filter-class>
</filter>

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

The code of the filter:

public class SummaryForwardFilter implements Filter{
    @Override
    public void init(FilterConfig filterConfig) throws ServletException {
    }

    @Override
    public void doFilter(ServletRequest servletRequest, ServletResponse servletResponse, FilterChain filterChain) throws IOException, ServletException {
        HttpServletRequest request = (HttpServletRequest)servletRequest;
        HttpServletResponse response = (HttpServletResponse)servletResponse;
        String url = request.getRequestURL().toString();
        String[] urlSegments = url.split("/");
        String guid = urlSegments[urlSegments.length-1];
        if(guid.matches("^[a-fA-F0-9]{8}(-[a-fA-F0-9]{4}){3}-[a-fA-F0-9]{12}$")){
            request.getRequestDispatcher("/WEB-INF/pages/summary.jsp").forward(servletRequest, servletResponse);
            return;
        }
        response.sendError(HttpServletResponse.SC_NOT_FOUND);
    }

    @Override
    public void destroy() {
    }
}

Then, the url http://mysite/mywebapp/summary/ecef22d6-7aa6-49db-b0d3-6577a63d14c8 will be mapped to the /WEB-INF/pages/summary.jsp. It's also okay when trying to map to non-jsp files.

The guid parameter can be retrieve by javascript code like that:

function extractGuid(value) {
    var re = /[0-9a-f]{8}-[0-9a-f]{4}-[1-5][0-9a-f]{3}-[89ab][0-9a-f]{3}-[0-9a-f]{12}/i;
    var match = re.exec(value);
    return match ? match[0] : null;
}

var guid = extractGuid(window.location.href);

Upvotes: 1

Leos Literak
Leos Literak

Reputation: 9474

Map your URL with default servlet (you do not need to declare it)

<servlet-mapping>
    <servlet-name>default</servlet-name>
    <url-pattern>/dosomething/*</url-pattern>
</servlet-mapping>

Upvotes: 1

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