Convert Spring MVC project to JAX-RS - spring

Question

I'm trying to do a project with JAX-RS, Spring, Hibernate, Maven and JPA. I'm following this article as reference. This project is in Spring MVC. I'm very new to Spring MVC. I have tried to convert it to JAX-RS. What are the important changes I have to do with this?
The main change I noticed that in Spring MVC we use like this: #RequestMapping("/accounts/{username}").
But in JAX-RS we use like this. #Path("/accounts/{username}")
Therefore If I do the changes like this, what are the other key changes we have to do with this?

Answer 1

The most basic thing you need to do in order to port the Spring MVC controllers to JAX-RS controllers, is to provide and integration between Spring and your JAX-RS provider. One way to do that is to make the appropriate changes to web.xml.
Assuming you are using Jersey 2.x and Spring 3.x, the web.xml would look like:
<web-app version="3.0" xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd">
<module-name>helloworld-spring</module-name>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:applicationContext.xml</param-value>
</context-param>
<servlet>
<servlet-name>SpringApplication</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>some.package.MyApplication</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>SpringApplication</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
Then MyApplication would look like:
package some.package;
import org.glassfish.jersey.server.ResourceConfig;
import org.glassfish.jersey.server.spring.scope.RequestContextFilter;
public class MyApplication extends ResourceConfig {
public MyApplication () {
register(RequestContextFilter.class);
register(JerseyResource.class);
register(SpringSingletonResource.class);
register(SpringRequestResource.class);
register(CustomExceptionMapper.class);
}
}
You will also need the dependency
<dependency>
<groupId>org.glassfish.jersey.ext</groupId>
<artifactId>jersey-spring3</artifactId>
<version>2.10.1</version>
</dependency>
besides the other Spring 3 and Jersey 2 dependencies.
Once you get that out of the way, you can start transforming the controllers.
For example:
#Path("admin")
#Component
public class AdminController {
#Autowired
private SchoolService service;
#Path("load-home-page")
#GET
public String loadHomePage(Map model){
return "admin/index";
}
//the rest of the code accordingly
}
The documentation for Spring-Jersey integration is here while a complete example can be found here.
Personally I would go the Spring MVC route instead of Jersey if you really don't have a compelling reason to choose Jersey over Spring MVC.