Hi,
For a past few days i have been trying to get ulc springmvc
integration working for a project. I could manage to get development
runner running as shown in one of the mails on the mailing list .
The problem arises when i access the application as a webapplication.
When i access the jsp page the SpringWebULCApplication gets
instantiated but when it goes to find the application using the
following
public static IApplication getApplication(ApplicationContext context,
String applicationBeanName) {
IApplication application;
if(applicationBeanName != null) {
application = ((IApplication) context.getBean(applicationBeanName));
} else {
final Map applications = context.getBeansOfType(IApplication.class);
if(applications.size() != 1) throw new
IllegalArgumentException("Application context must contain exactly
one ULCApplication: "+applications);
application = (IApplication)
applications.values().iterator().next();
}
return application;
}
it is not able to locate the bean which implements IApplication.class
and hence terminates with an error "Application context must contain
exactly one ULCApplication:"
Can someone help me in making use of SpringWebULCApplication class.
Is there anything more that is required like an
applicationContext.xml file to let the framework find the class that
implements IApplication.class ?
------------------------------------------------------------------------
So far i have managed to do the following changes.
1>Edit web.xml to introduce
<servlet>
<servlet-name>springapp</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>springapp</servlet-name>
<url-pattern>/ulc</url-pattern>
</servlet-mapping>
2>Introduce file called springapp-servlet.xml in the same location as
web.xml
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans";
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance";
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-2.5.xsd";>
<!-- the application context definition for the springapp
DispatcherServlet -->
<bean name="/ulc" class="org.mernst.ulcjava.spring.ULCController"/>
<bean
class="org.springframework.web.servlet.handler.BeanNameUrlHandlerMapping"/>
</beans>
3>Edit the xxxx.jsp file and update the url to access /ulc
<%
String applicationUrl = request.getScheme() + "://" +
request.getServerName() + ":" + request.getServerPort() +
request.getContextPath() + "/ulc";
%>
4>Introduce spring.xml in the resources ( i guess this would be only
used with development launcher code)
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans";
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance";
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-2.0.xsd";>
<bean name="ABC" scope="prototype" class="com.xxx.xxxx.application.ABC">
</bean>
</beans>
(Class ABC implements IApplication)
5>Development Runner Code
public class SpringDevelopmentRunner implements IApplication {
private static final String SPRING_XML = "/spring.xml";
private static final String SPRING_IAPPLICATION_BEAN = "ABC";
private ClassPathXmlApplicationContext _springContext;
private IApplication _delegate;
public void start() {
_springContext = new ClassPathXmlApplicationContext(SPRING_XML);
_delegate = (IApplication)
_springContext.getBean(SPRING_IAPPLICATION_BEAN);
_delegate.start();
}
...
.. ...
Thanks,
Eugene
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