Reputation: 38
Let me know how to initialize tiles from dispatcher and application context and if any changes required in web.xml, this is my tiles.xml
<definition name="contact" extends="mainLayout">
<put-attribute name="title" value="Contact Manager" />
<put-attribute name="body" value="/WEB-INF/jsp/contact.jsp" />
</definition>
</tiles-definitions>
in application context i have used
<bean id="tilesConfigurer" class="org.springframework.web.servlet.view.tiles2.TilesConfigurer">
<property name="definitions">
<list>
<value>/WEB-INF/tiles.xml</value>
</list>
</property>
</bean>
<bean id="viewResolver" class="org.springframework.web.servlet.view.UrlBasedViewResolver">
<property name="requestContextAttribute" value="requestContext"/>
<property name="viewClass" value="org.springframework.web.servlet.view.tiles2.TilesView"/>
</bean>
and dispatcher looks like this
<bean class="org.springframework.web.servlet.mvc.support.ControllerClassNameHandlerMapping"/>
indexController</prop>--> newController
<bean class="org.springframework.web.servlet.view.tiles2.TilesConfigurer" id="tilesConfigurer">
<property name="definitions">
<list>
<value>WEB-INF/tiles.xml</value>
</list>
</property>
</bean>
<bean class="org.springframework.web.servlet.view.UrlBasedViewResolver" id="viewResolver">
<property name="viewClass">
<value>
org.springframework.web.servlet.view.tiles2.TilesView
</value>
</property>
Upvotes: 0
Views: 494
Reputation: 48
In Dispatcher-Servlet
<bean id="tilesConfigurer" class="org.springframework.web.servlet.view.tiles2.TilesConfigurer">
<property name="definitions">
<list>
<value>/WEB-INF/tiles.xml</value>
</list>
</property>
</bean>
In web.xml use this code
<listener>
<listener-class>org.apache.tiles.web.startup.simple.SimpleTilesListener</listener-class>
</listener>
rest is not required for spring-tiles application, tiles.xml is depended on your requirement
Upvotes: 1