二、Servlet - 2.17 Servlet类开发步骤 - 《JavaWeb》

编写一个Servlet类，直接继承HttpServlet
重写doGet方法或者重写doPost方法，到底写谁？Javaweb程序员说了算 ```java package servlet;

import jakarta.servlet.ServletException; import jakarta.servlet.http.HttpServlet; import jakarta.servlet.http.HttpServletRequest; import jakarta.servlet.http.HttpServletResponse;

import java.io.IOException; import java.io.PrintWriter;

/**

@Author: 小雷学长
@Date: 2022/3/20 - 15:24

@Version: 1.8 */ public class LoginServlet extends HttpServlet { @Override protected void doPost(HttpServletRequest req, HttpServletResponse resp)

  throws ServletException, IOException {
  resp.setContentType("text/html");
  PrintWriter out = resp.getWriter();
  out.print("<h1>登录成功……</h1>");

} }


- [x] 将Servlet类配置到web.xml文件当中
```xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee"
         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_4_0.xsd"
         version="4.0">
    <servlet>
        <servlet-name>login</servlet-name>
        <servlet-class>servlet.LoginServlet</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>login</servlet-name>
        <url-pattern>/login</url-pattern>
    </servlet-mapping>
</web-app>

[x] 准备前端的页面（form表单），form表单中指定请求路径路径即可，

<!DOCTYPE html>
<html lang="en">
<head>
  <meta charset="UTF-8">
  <title>Login page</title>
</head>
<body>
  <form action="/Servlet00/login" method="post">
    <input type="submit" value="login">
  </form>
</body>
</html>