2008/6/27 Milan Milanovic <[EMAIL PROTECTED]>: > > Hi Lukasz, > > AFAIK I just need to define sessionFactory bean in my applicationContext.xml > to work with my JPA-enabled classes (just like in the Person example) ? How > can I do that ?
Sorry!! My big mistake ;-) Right now I discovered that you use JPA, not Hibernate. In such example there is such configuration below, did you setup it? <?xml version="1.0" encoding="UTF-8"?> <web-app id="person" version="2.4" xmlns="http://java.sun.com/xml/ns/j2ee"; xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"; xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd";> <display-name>person</display-name> <!-- Include this if you are using Hibernate --> <filter> <filter-name>Spring OpenEntityManagerInViewFilter</filter-name> <filter-class> org.springframework.orm.jpa.support.OpenEntityManagerInViewFilter </filter-class> </filter> <filter-mapping> <filter-name>Spring OpenEntityManagerInViewFilter</filter-name> <url-pattern>/*</url-pattern> </filter-mapping> <filter> <filter-name>struts2</filter-name> <filter-class> org.apache.struts2.dispatcher.FilterDispatcher </filter-class> </filter> <filter-mapping> <filter-name>struts2</filter-name> <url-pattern>/*</url-pattern> </filter-mapping> <welcome-file-list> <welcome-file>index.jsp</welcome-file> </welcome-file-list> <listener> <listener-class> org.springframework.web.context.ContextLoaderListener </listener-class> </listener> </web-app> Regards -- Lukasz http://www.lenart.org.pl/ --------------------------------------------------------------------- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
