用Jersey构建RESTful服务

一、环境

1.Eclipse Juno R2
2.Tomcat 7
3.Jersey 2.x(最新2.11版本测试通过) 下载地址（ https://jersey.java.net/download.html）

二、流程

1.Eclipse 中创建一个 Dynamic Web Project ,本例为“RestDemo”
2.按个各人习惯建好包，本例为“com.waylau.rest.resources”
3.解压jaxrs-ri-2.7，将api、ext、lib文件夹下的jar包都放到项目的lib下；项目引入jar包
4.在resources包下建一个class“HelloResource”

package com.waylau.rest.resources; 
 import javax.ws.rs.GET; 
 import javax.ws.rs.Path; 
 import javax.ws.rs.Produces; 
 import javax.ws.rs.PathParam; 
 import javax.ws.rs.core.MediaType; 
 @Path("/hello") 
 public class HelloResource { 
 @GET @Produces(MediaType.TEXT_PLAIN) 
 public String sayHello() { 
 return "Hello World!" ; 
 }

 @GET @Path("/{param}") 
 @Produces("text/plain;charset=UTF-8") 
 public String sayHelloToUTF8(@PathParam("param") String username) 
 { 
 return "Hello " + username; 
 }

 }

5.修改web.xml,添加基于Servlet-的部署

<servlet> <servlet-name>Way REST Service</servlet-name> <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class> <init-param> <param-name>jersey.config.server.provider.packages</param-name> <param-value>com.waylau.rest.resources</param-value> </init-param> <load-on-startup>1</load-on-startup> </servlet> <servlet-mapping> <servlet-name>Way REST Service</servlet-name> <url-pattern>/rest/*</url-pattern> </servlet-mapping>