Question

我创建了web动态项目，上下文名称是“restJeresy”。并在我的本地使用websphere liberty服务器。

当我点击网址时 http://localhost:9080/restJeresy/TestRestService/callService显示错误

Error 404: Not Found

任何人都可以帮助我，我缺少什么？

这是我的代码，

package com.demo;

import javax.ws.rs.GET;
import javax.ws.rs.Path;

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

     @Path("/callService")
     @GET
    public String getMsg() {
        return "Hello World !! - Jersey 2";
    }
}

Web.xml文件

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd" id="WebApp_ID" version="3.1">
  <display-name>restJeresy</display-name>
  <welcome-file-list>
    <welcome-file>index.html</welcome-file>
    <welcome-file>index.htm</welcome-file>
    <welcome-file>index.jsp</welcome-file>
    <welcome-file>default.html</welcome-file>
    <welcome-file>default.htm</welcome-file>
    <welcome-file>default.jsp</welcome-file>
  </welcome-file-list>
  <servlet>
      <servlet-name>Jersey RESTful Application</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.demo</param-value>
         </init-param>
         <load-on-startup>1</load-on-startup>
      </servlet>
   <servlet-mapping>
   <servlet-name>Jersey RESTful Application</servlet-name>
      <url-pattern>/</url-pattern>
   </servlet-mapping>  
</web-app>

Answer 1

web.xml中servlet映射的问题。请检查

server "default" {
    listen on 0.0.0.0 port 80

    fastcgi socket ":3031"
}

然后尝试访问http://localhost:9080/restJersey/restf/TestRestService/callService

错误404：Not Found错误显示Jeresy restful service

1 个答案: