带有tomcat的球衣给404错误

问题描述

我正在使用带有tomacat 6和Eclipse Juno的Jersey.

I am using Jersey with tomacat 6 and eclipse Juno.

我的文件名为HelloWorldService:

my file name is HelloWorldService :

package com.mkyong.rest;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.PathParam;
import javax.ws.rs.core.Response;

@Path("/hello")
public class HelloWorldService {

@GET
@Path("/{param}")
public Response getMsg(@PathParam("param") String msg) {

    String output = "Jersey say : " + msg;

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

  }

}

Web.xml:

<web-app id="WebApp_ID" version="2.4" xmlns="http://java.sun.com/xml/ns/j2ee"  xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee  http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">

<display-name>Restful Web Application</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.mkyong.rest</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

<servlet-mapping>
    <servlet-name>jersey-serlvet</servlet-name>
    <url-pattern>/rest/*</url-pattern>
</servlet-mapping>

当我使用Apache作为服务器运行我的项目时，它将转到此URL:

when i am running my project using Apache as server- it is going to this url:

http://localhost:8080/Test/WEB-INF/classes/com/mkyong/rest/HelloWorldService.java

我收到404错误.

我还尝试了version ="3.0"，但没有成功.请帮帮我.

I also tried with version="3.0" and it didn't work.Please help me out.

推荐答案

您应按如下所示提供您的网址；

you should give your url as follows;

http://localhost:8080/<your-war-name>/rest/hello/<your-param>

这篇关于带有tomcat的球衣给404错误的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！