I'm trying to integrate Spring 3.0.5 with Jersey 1.4. I seem to have everything working, but whenever I try and return a Viewable that points to a JSP, I get a 404 error. When I wasn't using spring I could use this filter:
<filter>
<filter-name>Jersey Filter</filter-name>
<filter-class>com.sun.jersey.spi.container.
servlet.ServletContainer</filter-class>
<init-param>
<param-name>com.sun.jersey.config.feature.Redirect</param-name>
<param-value>true</param-value>
</init-param>
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>cheetah.frontend.controllers</param-value>
</init-param>
<init-param>
<param-name>com.sun.jersey.config.feature.FilterForwardOn404</param-name>
<param-value>true</param-value>
</init-param>
<init-param>
<param-name>com.sun.jersey.config.property.WebPageContentRegex</param-name>
<param-value>/(images|css|jsp)/.*</param-value>
</init-param>
</filter>
And I could return a Viewable to any JSP's, image, css that were stored in the appropriate folder. However, now that I have to use the SpringServlet to get spring integration, I'm at a loss for how to access resources, since I can't use the
filter above. I've tried using this
servlet mapping with no luck:
<servlet>
<servlet-name>jerseyspring</servlet-name>
<servlet-class>com.sun.jersey.spi.spring.container.
servlet.SpringServlet</servlet-class>
<load-on-startup>1</load-on-startup>
<init-param>
<param-name>com.sun.jersey.config.property.WebPageContentRegex</param-name>
<param-value>/(images|css|jsp)/.*</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>jerseyspring</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
Does anyone know the proper configurations to achieve this?
Thanks for any help.
