用Jersey构建RESTful服务
作者:互联网
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>
- 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%E4%BD%A0%E5%A5%BD%E5%90%97,输出Hello Way你好吗
标签:ws,rs,rest,Hello,构建,import,RESTful,javax,Jersey 来源: https://www.cnblogs.com/tiancai/p/14285406.html