how to define contextParameter in Spring - spring

Question

I am working on a web project and I defined some properties in the context.xml of my Tomcat, like path, properties value for the configuration of the application.
My problem arrives when I want to write some JUnit tests, which are launched outside my web container, how can I define these parameters?
To be clear, in my context.xml (in Tomcat configuration directory), I have:
<Parameter name="myProperty" value="myValue" override="false"/>
And with Spring, I access it with:
<property name="property" value="#{myProperty}" />
But when I launch a junit test, the context.xml is not loaded, I need another way to define the property.
How can I do that?
To be more precise, the context.xml file which we are talking about is a file used by my Tomcat server, it does not follow the Spring schema and I think that I can't "import" it into Spring.
I already use the SpringJUnit4ClassRunner and the ContextConfiguration tag, it works fine, but now, I need to emulate/replace the Tomcat's behaviour to define this ContextParameters and retrieve my parameter...
I hope I am clearer :)

Answer 1

Try using something like this :
#RunWith(SpringJUnit4ClassRunner.class)
#ContextConfiguration("classpath:context.xml")
public class MyTestClass {
//put tests here
}
EDIT:
You can also specify a path to the context file :
#RunWith(SpringJUnit4ClassRunner.class)
#ContextConfiguration("file:src/main/resources/spring/config.xml")
public class MyTestClass {
//put tests here
}

Answer 2

For the suggestion above to work, you need to depend on the spring-test module.
You can also load your context file the good old fashion way.
ApplicationContext context = new ClassPathXmlApplicationContext("context.xml")
And then grab your bean by name.