这是我的html
form.html<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=gb2312" />
<title>Beer</title>
</head><body>
<h1 align="center">Beer Selection Page</h1>
<form method="post"
action="SelectBeer.do">
Select beer characteristics<p>
Color:
<select name="color" size="1">
<option>light</option>
<option>amber</option>
<option>brown</option>
<option>dark</option>
</select>
<br><br>
<center>
<input type="SUBMIT">
</center>
</body>
</html>
然后下面是我的xml文件
web.xml
<web-app 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/web-app_2_4.xsd"
version="2.4">
<servlet>
<servlet-name>Ch3 Beer</servlet-name>
<servlet-class>com.example.web.BeerSelect</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Ch3 Beer</servlet-name>
<url-pattern>/SelectBeer.do</url-pattern>
</servlet-mapping>
</web-app>
然后是我的servlet文件BeerSelect源码
package com.example.web;
import javax.servlet.*;
import javax.servlet.http.*;
import java.io.*;
public class BeerSelect extends HttpServlet {
public void doPost(HttpServletRequest request,HttpServletResponse response)
throws IOException,ServletException{
response.setContentType("text/html");
PrintWriter out=response.getWriter();
out.println("Beer Selection Advice<br>");
String c=request.getParameter("color");
out.println("<br>Got beer color"+c);
}
}
form.html放在F:\tomcat\apache-tomcat-8.5.20\webapps\Beer-v1目录下
web.xml放在F:\tomcat\apache-tomcat-8.5.20\webapps\Beer-v1\WBE-INF目录下
BeerSelect.class放在F:\tomcat\apache-tomcat-8.5.20\webapps\Beer-v1\WBE-INF\classes\com\example\web目录下,然后每次我打开form的文件,然后点击提交按钮,都会出现404错误,请问这是什么原因