How to register Spring @Configuration annotated class instead of applicationContext.xml file in web.xml?

Question

I am using jsf and spring together in web application. I have configured datasource and session factory in one configuration class which uses annotations like @Configuration, @ComponentScan etc. I don't have any applicationContext.xml file in my project as I am handling every entry of context xml in Configuration class. The test case works successfully but when I deploy my web application, it gives me error

java.lang.IllegalStateException: No WebApplicationContext found: no ContextLoaderListener registered?

Now if I give listener class in web.xml,

<listener> <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class> </listener>

it gives me error,

/WEB-INF/applicationContext.xml not found

As per the document of ContextLoaderListener, it's true that if I don't give contextConfigLocation param in web.xml explicitly, it will search for the default spring context file named applicationContext.xml in web.xml. Now, what should I do if I don't want to use spring context file and do all the configuration with annotations? How should I register listener class ContextLoaderListener so that without use of xml file and using annotations only, I be able to run my web application with spring and jsf?

Community · Accepted Answer · 2020-06-20 09:12:55Z

In web.xml you need to bootstrap the context with AnnotationConfigWebApplicationContext:

<servlet> <servlet-name>appServlet</servlet-name> <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class> <init-param> <param-name>contextClass</param-name> <param-value> org.springframework.web.context.support.AnnotationConfigWebApplicationContext </param-value> </init-param> <init-param> <param-name>contextConfigLocation</param-name> <param-value> org.package.YouConfigurationAnnotatedClass </param-value> </init-param> </servlet>

And don't forget to use @EnableWebMvc for your MVC annotations to kick in.

further reading:

EDIT as a "comments follow up" => to be Turing Complete:

Yes of course you need a listener. Although the above completely answers the question "How to register Spring @Configuration annotated class instead of applicationContext.xml file in web.xml", here is an example from Spring official documentation that layouts the full web.xml:

<web-app> <!-- Configure ContextLoaderListener to use AnnotationConfigWebApplicationContext instead of the default XmlWebApplicationContext --> <context-param> <param-name>contextClass</param-name> <param-value> org.springframework.web.context.support.AnnotationConfigWebApplicationContext </param-value> </context-param> <!-- Configuration locations must consist of one or more comma- or space-delimited fully-qualified @Configuration classes. Fully-qualified packages may also be specified for component-scanning --> <context-param> <param-name>contextConfigLocation</param-name> <param-value>com.acme.AppConfig</param-value> </context-param> <!-- Bootstrap the root application context as usual using ContextLoaderListener --> <listener> <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class> </listener> <!-- Declare a Spring MVC DispatcherServlet as usual --> <servlet> <servlet-name>dispatcher</servlet-name> <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class> <!-- Configure DispatcherServlet to use AnnotationConfigWebApplicationContext instead of the default XmlWebApplicationContext --> <init-param> <param-name>contextClass</param-name> <param-value> org.springframework.web.context.support.AnnotationConfigWebApplicationContext </param-value> </init-param> <!-- Again, config locations must consist of one or more comma- or space-delimited and fully-qualified @Configuration classes --> <init-param> <param-name>contextConfigLocation</param-name> <param-value>com.acme.web.MvcConfig</param-value> </init-param> </servlet> <!-- map all requests for /app/* to the dispatcher servlet --> <servlet-mapping> <servlet-name>dispatcher</servlet-name> <url-pattern>/app/*</url-pattern> </servlet-mapping> </web-app>

Hi, I am not using Spring MVC here, only JSF and Spring 3.1 RC1. Though I wrote this servlet in my web.xml but still got same error "no ContextLoaderListener registered". I then did entry for listener class of spring and then got other error of applicationContext.xml not found. I hope I am making my point clear.
Hey, I got it worked when I added these 2 params as context params and spring listener class in web.xml. Thanks, your answer helped me to find out my answer. I am editing your answer to add my solution. You please approve it (as I don't have edit privileges) so that I can mark your answer as final.
I never received you edit request, but I updated my answer with an example of a full web.xml form the Spring official documentation.
In my web application config i had to have <load-on-startup>1</load-on-startup> in dispatcher servlet. Without this dispatcher have not be loaded
@tolitius, thanks for this comment in your answer as it helped me understand the point

Angad · Accepted Answer · 2015-01-27 16:32:43Z

Bumping up an old question here, but with recent versions of Spring (v3.0+) now you can get rid of web.xml altogether, provided you are deploying your app on a web container that supports Servlet 3.0+.

One can implement Spring's WebApplicationInitializer interface to do the same configurations that one would do in web.xml. This implementation class will be automatically detected by Spring 3.0+ app running on Servlet 3.0+ containers.

If the set up is rather simple, you could instead use another class provided by Spring as shown below. All one does here is to set the @Configuration classes and list out the servlet mappings. Keeps the setup extremely simple.

public class WebInit extends AbstractAnnotationConfigDispatcherServletInitializer{ @Override protected Class<?>[] getRootConfigClasses() { return null; } @Override protected Class<?>[] getServletConfigClasses() { return new Class[] {AppConfig.class}; } @Override protected String[] getServletMappings() { return new String[] { "*.html" ,"*.json" ,"*.do"}; } }

If you do try to get rid of web.xml be sure you're using a container that supports Servlet 3.0 like Tomcat 7.0.57. P.S. Steve, nice work with the leading commas!

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