问答详情
源自:3-5 servlet类的实现

运行整个项目的时候点击生成报表就弹出以下错误,怎么解决?视频里面showReport创建的那一部分被剪掉了

HTTP Status 404 - /TestReport/ShowReport

type Status report

message /TestReport/ShowReport

description The requested resource is not available.



补充ShowReport代码

package servlet;

import java.io.IOException;

import java.util.List;

import javax.servlet.ServletException;

import javax.servlet.http.HttpServlet;

import javax.servlet.http.HttpServletRequest;

import javax.servlet.http.HttpServletResponse;

import service.Service;


public class ShowReport extends HttpServlet {

public ShowReport(){

super();

}


public void init() throws ServletException{

}


public void doGet(HttpServletRequest request,HttpServletResponse response) throws ServletException, IOException{

this.doPost(request, response);

}


public void doPost(HttpServletRequest request,HttpServletResponse response) throws IOException{

List list;

Service service = new Service();

list = service.getProfit();

request.getSession().setAttribute("New", list);

response.sendRedirect("NewFile.jsp");

}

}


提问者:qq_卡布奇诺_4 2015-06-16 15:29

个回答

  • ____途
    2018-02-26 14:20:52

    我也是

  • qq_双鱼座的猫_0
    2015-12-18 16:17:01

    你可以在web.xml中加上这两段,就可以了,这个配置在视频里老师可能是忘了说了。

        <!-- 根据名称找到控制层类 ShowReport.class--> 

      <servlet>

        <servlet-name>ShowReport</servlet-name>

        <servlet-class>servlet.ShowReport</servlet-class>

      </servlet>

       

      <!-- JSP 表单提交<form action="ShowReport" method="post"> --> 

      <servlet-mapping>

            <servlet-name>ShowReport</servlet-name>

            <url-pattern>/ShowReport</url-pattern>

      </servlet-mapping>


  • qq_卡布奇诺_4
    2015-06-17 10:59:12

    问题已解决,servlet3.0以后不需要在web.xml中添加代码,但是servlet类中需要以下类似声明

    @WebServlet(urlPatterns = {"/ShowReport"}, asyncSupported = true, 

    loadOnStartup = -1, name = "ShowReport", displayName = "sr"


  • 好帮手慕珊
    2015-06-17 09:17:29

    你是说ShowReport类创建的步骤没了,直接写代码了吗