Question

我遵循了本教程 http://www.javawebtutor.com/articles/web-services/rest/creating-web-service-with-maven-jersey.php

并使用类似这样的链接访问API

localhost:8080/RESTfullApp/rest/hello/inputhere

我需要使用以下链接替换现有的RESTFul API：

localhost:8080/Sample.pl?Task=GetStuff&StuffNumber=1

完成的事情：

@Path("/Sample.pl?")
public class HelloWorldService {

    @GET
    @Path("Task=GetStuff&StuffNumber={name}")
    public Response getMsg(@PathParam("name") String name) {

        String output = "Stuff num  : " + name;

        return Response.status(200).entity(output).build();

    }

}

web.xml：这里要改变什么？

<!DOCTYPE web-app PUBLIC
 "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
 "http://java.sun.com/dtd/web-app_2_3.dtd" >

<web-app>
    <display-name>RESTfulExample</display-name>
    <servlet>
        <servlet-name>jersey-serlvet</servlet-name>
        <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
        <init-param>
             <param-name>com.sun.jersey.config.property.packages</param-name>
             <param-value>com.jwt.rest</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>jersey-serlvet</servlet-name>
        <url-pattern>/*I dont know what to do here*/</url-pattern>
    </servlet-mapping>

</web-app>

如何更改Jax-RS + Jersey URL

0 个答案: