tomcat 部署 RESTful 服务实例

　　1、建立简单restfule服务

　　　　简单代码：

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)
    //http://localhost:8080/api/myresource/getIt
    @Path("/getIt")

    //localhost:8080/api/myresource/getIt
    public String getIt() {
        return "Got it!";
    }
}

　　　　web.xml文件

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee"
         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd"
         version="3.1">
    <servlet>
        <servlet-name>test</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.example</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>test</servlet-name>
        <url-pattern>/api/*</url-pattern>
    </servlet-mapping>
</web-app>