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Spring Framework - How to get Application Context when Spring Framework is started from Tomcat Serve

 
Greenhorn
Posts: 3
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I am a beginner is Spring Framework.
1. Today I have a non Web Application which uses Spring Framework.
I have a class MainApp. Inside main of that class, I have following code segment:
ApplicationContext context =
new ClassPathXmlApplicationContext("Beans.xml");
TWDBJDBCTemplate twdbJDBCTemplate =
(TWDBJDBCTemplate)context.getBean("twdbJDBCTemplate");
INTFDBJDBCTemplate intfJDBCTemplate =
(INTFDBJDBCTemplate)context.getBean("intfdbJDBCTemplate");
Where beans.xml defines various beans for Sprig Framework ..

2. Now I am developing a webservice app using Eclipse Dynamic Web Project.. I do not use Spring Framework webservices at this point since that requires ne learning.
I have modified Apache Tomcat web.xml file to load beans.xml file:
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/application-context.xml</param-value>
</context-param>

<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
When I start tomcat I do see beans.xml file loaded…
3. My question is inside my web service app Java classes, I need to get
ApplicationContext context = ???
So I can use this context, to get beans defined in web.xml file, e.g.,
TWDBJDBCTemplate twdbJDBCTemplate =
(TWDBJDBCTemplate)context.getBean("twdbJDBCTemplate");
Appreciate if you can provide guidance how to do that. Thanks a lot,


 
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May save you some pain if you look at using Spring Boot, in particular the Spring Boot Web starter project.
Start here - https://spring.io/guides/gs/spring-boot/

 
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