運(yùn)行整個(gè)項(xiàng)目的時(shí)候點(diǎn)擊生成報(bào)表就彈出以下錯(cuò)誤，怎么解決？視頻里面showReport創(chuàng)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.
補(bǔ)充ShowReport代碼
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");
}
}
2018-02-26
我也是
2015-12-18
你可以在web.xml中加上這兩段，就可以了，這個(gè)配置在視頻里老師可能是忘了說(shuō)了。
? ? <!-- 根據(jù)名稱找到控制層類 ShowReport.class-->?
? <servlet>
? ? <servlet-name>ShowReport</servlet-name>
? ? <servlet-class>servlet.ShowReport</servlet-class>
? </servlet>
? ?
? <!-- JSP 表單提交<form action="ShowReport" method="post"> -->?
? <servlet-mapping>
? ? ? ? <servlet-name>ShowReport</servlet-name>
? ? ? ? <url-pattern>/ShowReport</url-pattern>
? </servlet-mapping>
2015-06-17
問(wèn)題已解決，servlet3.0以后不需要在web.xml中添加代碼，但是servlet類中需要以下類似聲明
@WebServlet(urlPatterns = {"/ShowReport"}, asyncSupported = true,?
loadOnStartup = -1, name = "ShowReport", displayName = "sr"
)?
2015-06-17
你是說(shuō)ShowReport類創(chuàng)建的步驟沒(méi)了，直接寫代碼了嗎