How do I have common error page templates with tiles in a Spring/MVC 3.0 app?

Brett Ryan picture Brett Ryan · Aug 24, 2010 · Viewed 9.7k times · Source

I have a Spring MVC/3.0 app using tiles as it's view, this is working fine however I can't figure out how to get the error pages to also use tiles.

I have in my web.xml

<error-page>
  <error-code>404</error-code>
  <location>/WEB-INF/error/404.jsp</location>
</error-page>

which works fine as an ordinary view NOT using tiles, however when I change the location to one of the view names, the view is not found and renders the ordinary error page.

My tiles.xml file for the view contains the following definition

<definition name="404" extends="standardLayout">
  <put-attribute name="body" value="/WEB-INF/error/404.jsp" />
</definition>

I'm configuring tiles through spring as follows:

<bean id="tilesConfigurer" class="org.springframework.web.servlet.view.tiles2.TilesConfigurer">
  <property name="definitions">
    <list>
      <value>/WEB-INF/**/tiles.xml</value>
    </list>
  </property>
</bean>

I'm suspecting this is all due to the view not coming from spring itself?

Answer

naikus picture naikus · Aug 24, 2010

You need to add the "layouted" jsp in your web.xml. Below is the explaination code:

// Your web.xml should look like this:
<error-page>
  <error-code>404</error-code>
  <location>/WEB-INF/error/layout-404.jsp</location>
</error-page>


// Your layout-404.jsp should look like this:
<%@page isELIgnored="false" %>
<%@page contentType="text/html"%>
<%@taglib uri="http://tiles.apache.org/tags-tiles" prefix="tiles" %>
<tiles:insertDefinition name="404" />    


// Your layout def should look like this:
<definition name="404" extends="standardLayout">
  <put-attribute name="body" value="/WEB-INF/error/404.jsp" />
</definition>