
Originally Posted by
teagan05
In WEB.XML change the servlet mapping giving the url pattern.
Thanks - but that doesn't answer the question. Given the following defn:
Code:
<servlet>
<servlet-name>localization.xml.LocalizationServlet</servlet-name>
<servlet-class>org.springframework.web.context.support.HttpRequestHandlerServlet</servlet-class>
<init-param>
<param-name>debug</param-name>
<param-value>0</param-value>
</init-param>
<init-param>
<param-name>listings</param-name>
<param-value>true</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>localization.xml.LocalizationServlet</servlet-name>
<url-pattern>/localization/*</url-pattern>
</servlet-mapping>
How do I get my LocalizationServlet class to access the values of debug and listings?
Thanks,
Eric