Note
Not much comments on this official RESTEasy guide to integrate Spring, the documentation need to be improve, because it’s simply too “abstract” for others to understand (at least to me :)).
Here we show you two general ways to inject Spring bean into JBoss RESTEasy, below solutions should works on most of the web frameworks and JAX-RS implementations also.
- Use WebApplicationContextUtils + ServletContext.
- Create a class to implement ApplicationContextAware interfaces, with singleton pattern is recommended.
In this tutorial, we are integrating RESTEasy 2.2.1.GA with Spring 3.0.5.RELEASE.
1. Project Dependencies
Based on the article: RESTEasy hello world example with the following Maven command to create web project
mvn archetype:generate -DgroupId=com.tuts.heomi.netmon -DartifactId=resteasy-spring-integration-example -DarchetypeArtifactId=maven-archetype-webapp -DinteractiveMode=falseConvert to Eclipse web project:
mvn eclipse:eclipse -Dwtpversion=2.0Not many dependencies, just declare RESTEasy and Spring core in your Maven pom.xml file.
<repositories> <repository> <id>JBoss repository</id> <url>https://repository.jboss.org/nexus/content/groups/public-jboss/</url> </repository> </repositories> <dependencies> <!-- JBoss RESTEasy --> <dependency> <groupId>org.jboss.resteasy</groupId> <artifactId>resteasy-jaxrs</artifactId> <version>2.2.1.GA</version> </dependency> <!-- Spring 3 dependencies --> <dependency> <groupId>org.springframework</groupId> <artifactId>spring-core</artifactId> <version>3.0.5.RELEASE</version> </dependency> <dependency> <groupId>org.springframework</groupId> <artifactId>spring-context</artifactId> <version>3.0.5.RELEASE</version> </dependency> <dependency> <groupId>org.springframework</groupId> <artifactId>spring-web</artifactId> <version>3.0.5.RELEASE</version> </dependency> <!-- need for solution 1, if your container don't have this --> <dependency> <groupId>javax.servlet</groupId> <artifactId>servlet-api</artifactId> <version>2.4</version> </dependency> </dependencies>
2. Spring Bean
A simple Spring bean named “customerBo“, later you will inject this bean into RESTEasy service, via Spring.
package com.favtuts.customer;
public interface CustomerBo{
String getMsg();
}
package com.favtuts.customer.impl;
import com.favtuts.customer.CustomerBo;
public class CustomerBoImpl implements CustomerBo {
public String getMsg() {
return "RESTEasy + Spring example";
}
}
File : applicationContext.xml
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd ">
<bean id="customerBo" class="com.favtuts.customer.impl.CustomerBoImpl">
</bean>
</beans>
3.1 WebApplicationContextUtils + ServletContext
The first solution is uses JAX-RS @Context to get the ServletContext, and WebApplicationContextUtils to get the Spring application context, with this Spring application context, you are able to access and get beans from Spring container.
package com.favtuts.rest;
import javax.servlet.ServletContext;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.core.Context;
import javax.ws.rs.core.Response;
import org.springframework.context.ApplicationContext;
import org.springframework.web.context.support.WebApplicationContextUtils;
import com.favtuts.customer.CustomerBo;
@Path("/customer")
public class PrintService {
CustomerBo customerBo;
@GET
@Path("/print1")
public Response printMessage(@Context ServletContext servletContext) {
//get Spring application context
ApplicationContext ctx =
WebApplicationContextUtils.getWebApplicationContext(servletContext);
customerBo= ctx.getBean("customerBo",CustomerBo.class);
String result = customerBo.getMsg();
return Response.status(200).entity(result).build();
}
}
3.2 Implement ApplicationContextAware
The second solution is create a class implement Spring ApplicationContextAware, and make it singleton, to prevent instantiation from other classes.
package com.favtuts.context;
import org.springframework.beans.BeansException;
import org.springframework.context.ApplicationContext;
import org.springframework.context.ApplicationContextAware;
public class SpringApplicationContext implements ApplicationContextAware {
private static ApplicationContext appContext;
// Private constructor prevents instantiation from other classes
private SpringApplicationContext() {}
@Override
public void setApplicationContext(ApplicationContext applicationContext)
throws BeansException {
appContext = applicationContext;
}
public static Object getBean(String beanName) {
return appContext.getBean(beanName);
}
}
Remember register this bean, otherwise, Spring container will not aware of this class.
File : applicationContext.xml
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd ">
<bean class="com.favtuts.context.SpringApplicationContext"></bean>
<bean id="customerBo" class="com.favtuts.customer.impl.CustomerBoImpl">
</bean>
</beans>
In REST service, you can use the new singleton class – “SpringApplicationContext“, to get bean from Spring container.
package com.favtuts.rest;
import javax.servlet.ServletContext;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.core.Context;
import javax.ws.rs.core.Response;
import org.springframework.context.ApplicationContext;
import org.springframework.web.context.support.WebApplicationContextUtils;
import com.favtuts.context.SpringApplicationContext;
import com.favtuts.customer.CustomerBo;
@Path("/customer")
public class PrintService {
CustomerBo customerBo;
@GET
@Path("/print1")
public Response printMessage(@Context ServletContext servletContext) {
// get Spring application context
ApplicationContext ctx = WebApplicationContextUtils.getWebApplicationContext(servletContext);
customerBo = ctx.getBean("customerBo", CustomerBo.class);
String result = customerBo.getMsg();
return Response.status(200).entity(result).build();
}
@GET
@Path("/print2")
public Response printMessage() {
customerBo = (CustomerBo) SpringApplicationContext.getBean("customerBo");
String result = customerBo.getMsg();
return Response.status(200).entity(result).build();
}
}
4. Integrate RESTEasy with Spring
To integrate both in a web application, add Spring “ContextLoaderListener” in your web.xml.
File :web.xml
<web-app id="WebApp_ID" version="2.4"
xmlns="http://java.sun.com/xml/ns/j2ee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee
http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<display-name>Restful Web Application</display-name>
<context-param>
<param-name>resteasy.resources</param-name>
<param-value>com.favtuts.rest.PrintService</param-value>
</context-param>
<listener>
<listener-class>
org.jboss.resteasy.plugins.server.servlet.ResteasyBootstrap
</listener-class>
</listener>
<listener>
<listener-class>
org.springframework.web.context.ContextLoaderListener
</listener-class>
</listener>
<servlet>
<servlet-name>resteasy-servlet</servlet-name>
<servlet-class>
org.jboss.resteasy.plugins.server.servlet.HttpServletDispatcher
</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>resteasy-servlet</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
5. Demo
Both solution 1 and 2 will generate following same output :
Note
Again, if you understand this RESTEasy Spring guide, or has a better solution, please share it to me.
Download Source Code
$ git clone https://github.com/favtuts/java-jax-rs-tutorials.git
$ cd resteasy-spring-integrationRun web project on JBoss EAP 7.4
Access:
http://localhost:8080/resteasy-spring-integration-example/customer/print1
http://localhost:8080/resteasy-spring-integration-example/customer/print2