How to configure Spring MVC with pure Java-based configuration?

Question

I have, what I would consider a pretty simple Spring MVC setup. My applicationContext.xml is this:

<mvc:annotation-driven /> <mvc:resources mapping="/css/**" location="/css/" /> <context:property-placeholder location="classpath:controller-test.properties" /> <bean class="org.springframework.web.servlet.view.InternalResourceViewResolver"     p:prefix="/WEB-INF/views/" p:suffix=".jsp" /> 

My web.xml is currently this:

  <servlet>    <servlet-name>springDispatcherServlet</servlet-name>    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>    <init-param>       <param-name>contextConfigLocation</param-name>       <param-value>classpath:applicationContext.xml</param-value>    </init-param>    <load-on-startup>1</load-on-startup>   </servlet>    <!-- Map all requests to the DispatcherServlet for handling -->   <servlet-mapping>     <servlet-name>springDispatcherServlet</servlet-name>     <url-pattern>/</url-pattern>   </servlet-mapping> 

I am trying to convert this set up to pure Java-based config. I've searched the web and so far, I've come up with stuff that explains (some what) how to do the Java config but doesn't explain how to register that Java config with the environment, i.e., the web context.

What I have so far in terms of @Configuration is this:

 @Configuration  @EnableWebMvc  @PropertySource("classpath:controller.properties")  @ComponentScan("com.project.web")  public class WebSpringConfig extends WebMvcConfigurerAdapter {   @Override  public void addResourceHandlers(ResourceHandlerRegistry registry) {     registry.addResourceHandler("/css/**").addResourceLocations("/css/");  }   @Bean  public ViewResolver configureViewResolver() {      InternalResourceViewResolver viewResolve = new InternalResourceViewResolver();      viewResolve.setPrefix("/WEB-INF/views/");      viewResolve.setSuffix(".jsp");       return viewResolve;  }   @Override  public void configureDefaultServletHandling(DefaultServletHandlerConfigurer configurer){    configurer.enable();  } } 

How do I register this with the web container? I am using the latest spring (4.02).

Thanks!

Answer 1

You need to make following changes to web.xml in order to support java based configuration. This will tell the the DispatcherServlet to load configuration using the annotation based java configuration AnnotationConfigWebApplicationContext. You only need to pass the location of your java config file to the contextConfigLocation param, as below

<servlet>   <servlet-name>springDispatcherServlet</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>/*path to your WebSpringConfig*/ </param-value>   </init-param>   <load-on-startup>1</load-on-startup> </servlet> 

Update: Doing the same without making changes to web.xml

You can even do this without web.xml as Servlet Specification 3.0 makes the web.xml optional. You only need to implement/configure WebApplicationInitializer interface to configure the ServletContext which will allow you to create, configure, and perform registration of DispatcherServlet programmatically. The good thing is that WebApplicationInitializer is detected automatically.

In summary, one needs to implement WebApplicationInitializer to get rid of web.xml.

 public class MyWebAppInitializer implements WebApplicationInitializer {   @Override  public void onStartup(ServletContext container) {   // Create the 'root' Spring application context   AnnotationConfigWebApplicationContext rootContext =                        new AnnotationConfigWebApplicationContext();   rootContext.register(WebSpringConfig.class);    // Manage the lifecycle of the root application context   container.addListener(new ContextLoaderListener(rootContext));    // Create the dispatcher servlet's Spring application context   AnnotationConfigWebApplicationContext dispatcherContext =                      new AnnotationConfigWebApplicationContext();   dispatcherContext.register(DispatcherConfig.class);    // Register and map the dispatcher servlet   ServletRegistration.Dynamic dispatcher =     container.addServlet("dispatcher", new DispatcherServlet(dispatcherContext));     dispatcher.setLoadOnStartup(1);     dispatcher.addMapping("/");   } } 

Update: from comments
A slightly more convoluted explanation is also included in the official Spring reference Spring 4 Release

Reference:

http://docs.spring.io/spring/docs/3.1.x/javadoc-api/org/springframework/web/WebApplicationInitializer.html