用Jersey构建RESTful服务

一：环境

1、Tomcat
7
2. Jersey 2.26 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“RestfulDemo”


一路next到第三步时记得勾选



2、新建一个包



4.将下载的jesrey的文件全部放在lib文件下



5、在resources包下建一个class“HelloResource”

package com.chenll.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;
}

}

6、修改web.xml,添加基于Servlet-的部署

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0">
<display-name>RestfulDemo</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
<welcome-file>default.html</welcome-file>
<welcome-file>default.htm</welcome-file>
<welcome-file>default.jsp</welcome-file>
</welcome-file-list>
<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.chenll.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>
</web-app>

7.项目部署到tomcat,运行
8.浏览器输入要访问的uri地址 http://localhost:8088/RestfulDemo/rest/hello http://localhost:8088/RestfulDemo/rest/hello/l浪潮