创建一个新项目：

使用maven  & Eclipse (Mars)

Maven命令行创建一个新项目（maven环境搭建如不知晓自己查去）

mvn archetype:generate -DarchetypeArtifactId=jersey-quickstart-grizzly2 -DarchetypeGroupId=org.glassfish.jersey.archetypes -DinteractiveMode=false -DgroupId=com.example -DartifactId=simple-service -Dpackage=com.example -DarchetypeVersion=2.14

这要下载一些包 耐心等候……

处理完毕后，你会发现simple-service已经创建

导入到eclipse中,(eclipse要安装maven插件，怎么安装，自己去查)

导入成功后的目录结构如下图

打开包com.example下的Main.java

package com.example;

import org.glassfish.grizzly.http.server.HttpServer;
import org.glassfish.jersey.grizzly2.httpserver.GrizzlyHttpServerFactory;
import org.glassfish.jersey.server.ResourceConfig;

import java.io.IOException;
import java.net.URI;

/**
 * Main class.
 *
 */
public class Main {
    // Base URI the Grizzly HTTP server will listen on
    public static final String BASE_URI = "http://localhost:8080/myapp/";

    /**
     * Starts Grizzly HTTP server exposing JAX-RS resources defined in this application.
     * @return Grizzly HTTP server.
     */
    public static HttpServer startServer() {
        // create a resource config that scans for JAX-RS resources and providers
        // in com.example package
        final ResourceConfig rc = new ResourceConfig().packages("com.example");

        // create and start a new instance of grizzly http server
        // exposing the Jersey application at BASE_URI
        return GrizzlyHttpServerFactory.createHttpServer(URI.create(BASE_URI), rc);
    }

    /**
     * Main method.
     * @param args
     * @throws IOException
     */
    public static void main(String[] args) throws IOException {
        final HttpServer server = startServer();
        System.out.println(String.format("Jersey app started with WADL available at "
                + "%sapplication.wadl\nHit enter to stop it...", BASE_URI));
        System.in.read();
        server.stop();
    }
}

运行之后，控制台输出如下信息：

按照提示我们在浏览器输入http://localhost:8080/myapp/application.wadl，奇迹出现了，如下图所示(⊙o⊙)…

下一个提示出现了 http://localhost:8080/myapp/myresource,返回了一串文本信息。如下图

为什么会是这个访问路径，又为什么返回Got it ! 到底是怎么回事呢。我啥都不知道啊！

com.example还有个文件,MyResource.java,打开看看。

package com.example;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;

/**
 * Root resource (exposed at "myresource" path)
 */
@Path("myresource")
public class MyResource {

    /**
     * Method handling HTTP GET requests. The returned object will be sent
     * to the client as "text/plain" media type.
     *
     * @return String that will be returned as a text/plain response.
     */
    @GET
    @Produces(MediaType.TEXT_PLAIN)
    public String getIt() {
        return "Got it!";
    }
}

命令行那句话，Hit enter to stop it...  OK,stop it.

mission completed！
O(∩_∩)O~

JAVA RESTful Web Services - Jersey 入门