Tiles Integration in Spring 3

Hello to all,

I have create a web site using Tiles for templating purposes but when i redirect the index.jsp to home.jsp. It throw exception unable to resolve the view name. There is home.jsp available in WebContent folder.

Why is it cannot resolve the view name ?
Is there any method to resolve the view name to default if nothing found?

All my jsp files stores in WebContent folder. All my xml files stores in WEB-INF folder.

Tiles-def.xml

Code:

<tiles-definitions> <definition name="masterTemplate" template="/WebContent/layout.jsp"> <put-attribute name="header" value="/WebContent/header.jsp"></put-attribute> <put-attribute name="menu" value="/WebContent/menu.jsp"></put-attribute> <put-attribute name="body" value=""></put-attribute> <put-attribute name="footer" value="/WebContent/footer.jsp"></put-attribute> </definition> <definition name="home" extends="masterTemplate"> <put-attribute name="body" value="/WebContent/home.jsp"></put-attribute> </definition> </tiles-definitions>

web.xml

Code:

<display-name>Spring_Tiles</display-name> <welcome-file-list> <welcome-file>index.jsp</welcome-file> </welcome-file-list> <listener> <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class> </listener> <servlet> <servlet-name>dispatcher</servlet-name> <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class> <init-param> <param-name>contextConfigLocation</param-name> <param-value>/WEB-INF/applicationContext.xml</param-value> </init-param> <load-on-startup>1</load-on-startup> </servlet> <servlet-mapping> <servlet-name>dispatcher</servlet-name> <url-pattern>/</url-pattern> </servlet-mapping>

dispatcher-servlet.xml

Code:

<context:component-scan base-package="com.peter.controller"></context:component-scan>      <bean id="viewResolver" class="org.springframework.web.servlet.view.UrlBasedViewResolver"> <property name="viewClass" value="org.springframework.web.servlet.view.tiles2.TilesView"></property> </bean>  <bean id="tilesConfigurer" class="org.springframework.web.servlet.view.tiles2.TilesConfigurer"> <property name="definitions"> <list> <value>/WEB-INF/tiles-def.xml</value> </list> </property> </bean> <mvc:annotation-driven />

applicationContext.xml

Code:

 <import resource="dispatcher-servlet.xml"></import>

index.jsp

Code:

<body> Spring Tiles <jsp:forward page="/Spring_Tiles/showHome"></jsp:forward> </body>

HomeController.java

Code:

@Controller public class HomeController { /** * */ public HomeController() { } @RequestMapping(value = "/showHome", method = RequestMethod.GET) public String showHome() { return "/home"; } }

Please help.

Thanks.

StackTrace:
Could not resolve view with name /name in servlet with name 'dispatcher'

org.springframework.web.servlet.Dispatcher.render

I suggest you take a look at your configuration and code... The name of your view is home not /home...

Your application is not configured correctly. You should have one application context that contains beans that are "universal" to your application, while you can have many dispatcher servlets configured. Think about it this way -- your context should know about the application, but not about the implementation details (the servlets), while your servlets should have knowledge of the application, but not each other. It should look something like this:

web.xml

Code:

<?xml version="1.0" encoding="UTF-8"?> <web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"> <display-name>your-app</display-name> <context-param> <param-name>contextConfigLocation</param-name> <param-value>/WEB-INF/applicationContext.xml</param-value> </context-param> <listener> <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class> </listener> <servlet> <servlet-name>spring</servlet-name> <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class> <load-on-startup>1</load-on-startup> </servlet> <servlet-mapping> <servlet-name>spring</servlet-name> <url-pattern>/</url-pattern> </servlet-mapping> </web-app>

Now create two configuration files, one called applicationContext.xml, and the other called spring-servlet.xml, and place them both under the /WEB-INF/ directory. Actually, if you don't want to rename applicationContext.xml to something different, then you can do away with the contextConfigLocation context parameter altogether. Also, don't bother with the index.jsp redirection, and remove the welcome file list. You can create a controller that catches the "root" or "home" with the above configuration and the following handler:

HomeController.java handler:

Code:

@RequestMapping(value={"/",""}) public String index() throws Exception { return "home"; }

Finally, for your tiles configuration, as was already mentioned, your tiles definition name needs to match exactly the string you return from your handler method (so "home").

By using InternalResourceViewResolver, i will specify "/home" but when using URIBasedView Resolver, need to specify "home"(View name) and not the URI itself.

For all the view in jsp, need to specify in Tiles configuration file, it is quite bad.
Why it is view name rather that it is URL?

Is this correct?

I found one thing very strange where view cannot be render (display header.jsp, menu.jsp blah blah) when jsp file locate at WebContent folder but when i move to WEB-INF folder, the view can be render. What is the reason?

Thanks.