Servlet类开发步骤

  • 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
  1. package servlet;
  3. import jakarta.servlet.ServletException;
  4. import jakarta.servlet.http.HttpServlet;
  5. import jakarta.servlet.http.HttpServletRequest;
  6. import jakarta.servlet.http.HttpServletResponse;
  8. import java.io.IOException;
  9. import java.io.PrintWriter;
  11. /**
  12. * @Author: 小雷学长
  13. * @Date: 2022/3/20 - 15:24
  14. * @Version: 1.8
  15. */
  16. public class LoginServlet extends HttpServlet {
  17.     @Override
  18.     protected void doPost(HttpServletRequest req, HttpServletResponse resp)
  19.         throws ServletException, IOException {
  21.         resp.setContentType("text/html");
  22.         PrintWriter out = resp.getWriter();
  23.         out.print("<h1>登录成功……</h1>");
  24.     }
  25. }
  • 将Servlet类配置到web.xml文件当中
  1. <?xml version="1.0" encoding="UTF-8"?>
  2. <web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee"
  3.          xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
  4.          xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_4_0.xsd"
  5.          version="4.0">
  7.     <servlet>
  8.         <servlet-name>login</servlet-name>
  9.         <servlet-class>servlet.LoginServlet</servlet-class>
  10.     </servlet>
  11.     <servlet-mapping>
  12.         <servlet-name>login</servlet-name>
  13.         <url-pattern>/login</url-pattern>
  14.     </servlet-mapping>
  15. </web-app>
  • 准备前端的页面（form表单），form表单中指定请求路径路径即可，
  1. <!DOCTYPE html>
  2. <html lang="en">
  3.   <head>
  4.     <meta charset="UTF-8">
  5.     <title>Login page</title>
  6.   </head>
  7.   <body>
  8.     <form action="/Servlet00/login" method="post">
  9.       <input type="submit" value="login">
  10.     </form>
  12.   </body>
  13. </html>

17、Servlet类开发步骤 - 图1

17、Servlet类开发步骤 - 图2