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
qq_卡布奇诺_4
2015-06-16 15:29
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");
}
}
4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
使用JSP+Servlet技术生成报表
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