用Jersey构建RESTful服务1--HelloWorld

一、环境
1、Eclipse Juno R2
2. Tomcat 7
3. Jersey 2.7  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>

6.项目部署到tomcat,运行
7.浏览器输入要访问的uri地址
http://localhost:8089/RestDemo/rest/hello

输出Hello World!

http://localhost:8089/RestDemo/rest/hello/Way你好吗


输出Hello Way你好吗

 

参考：https://jersey.java.net/documentation/latest/user-guide.html