从mysql数据库登录验证servlet

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【中文标题】从mysql数据库登录验证servlet【英文标题】:login validate servlet from mysql database 【发布时间】:2012-11-28 07:57:47 【问题描述】:

我尝试创建一个简单的登录表单,它采用用户名和密码,检查 mysql 数据库表。对不起,我是这个 java 东西的初学者......如果它匹配它重定向到主页。但我可以'不执行它。有人可以帮我解决这个问题。感谢您的快速回复。我得到的 tomcat 错误是请求的资源 (/UserDemo/firstserv) 不可用。我知道其中还有更多错误。就是你,我把它贴在这里....帮帮我....

SerExam.java

 package myPack;

import java.io.IOException;
import java.io.PrintWriter;
import java.sql.*;
import javax.servlet.RequestDispatcher;
import javax.servlet.ServletConfig;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class SerExam extends HttpServlet

      Connection con;
      PreparedStatement ps;
      ResultSet rs;
      public void init(ServletConfig config)throws ServletException
      
            try
               
                        Class.forName("oracle.jdbc.driver.OracleDriver");
                        con=DriverManager.getConnection("jdbc:oracle:thin:@localhost:1521:xe","system","tiger");
               
                  catch (ClassNotFoundException e)
                     
                        System.out.println(e);
                     
                  catch (SQLException e)
                     
                        System.out.println(e);
                     
      
      protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException
      
            doPost(request, response);
            
                response.setContentType("text/html");
                PrintWriter pw=response.getWriter();
                String username=request.getParameter("username");
                String password=request.getParameter("password");
                pw.println("<html><body>");
                try
                
                      ps=con.prepareStatement("select * from loginvalidation where username=? and password=?");
                      ps.setString(1, username);
                      ps.setString(2, password);
                      rs=ps.executeQuery();
                      if(rs.next())
                      
                            pw.println("<h3>welcome " +" " + username +"</h3>");
                            RequestDispatcher rd1=request.getRequestDispatcher("./home.html");
                            rd1.include(request,response);
                            //or
                            //response.sendRedirect("./home.html");
                            pw.println("<form method=\"post\" action=\"Login.html\">");
                            pw.println("<input type=\"submit\" name=\"logout\" " + "value=\"Logout\">");
                            pw.println("</form>");

                      
                      else
                      
                            pw.println("<center><h3>invalid username/password Enter Correct username/password</h3></center>");
                            RequestDispatcher rd2=request.getRequestDispatcher("./Login.html");
                            rd2.include(request,response);
                            //or
                            //response.sendRedirect("./Login.html");
                      
                
                catch (SQLException e)



    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException
    
    

登录.html

<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Login page</title>
</head>
<body>
<center>
<form action="./firstserv" method="post">
username &nbsp;&nbsp;&nbsp;
<input type="text" name="username" />
<br>
<br>
password &nbsp;&nbsp;&nbsp;
<input type="password" name="password"></input><br><br>
&nbsp;&nbsp;&nbsp;
<input type="submit" value="login"></input>
&nbsp;&nbsp;&nbsp;
<a href="./reg.html">new user</a>
</form>
</center>
</body>
</html>

reg.html

<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Registration Form</title>
</head>
<body>
registration page under construction...............

</body>
</html>

web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" version="2.4" 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 http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
    <display-name>UserDemo</display-name>
    <servlet>
        <description>
        </description>
        <display-name>SerExam</display-name>
        <servlet-name>SerExam</servlet-name>
        <servlet-class>
        myPack.SerExam</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>SerExam</servlet-name>
        <url-pattern>/home.html</url-pattern>
    </servlet-mapping>
    <welcome-file-list>
        <welcome-file>index.html</welcome-file>
        <welcome-file>index.htm</welcome-file>
        <welcome-file>index.jsp</welcome-file>
        <welcome-file>default.html</welcome-file>
        <welcome-file>default.htm</welcome-file>
        <welcome-file>default.jsp</welcome-file>
    </welcome-file-list>
</web-app>

【问题讨论】:

【参考方案1】:

将您的 servlet 映射标记更改为

<servlet-mapping>
    <servlet-name>SerExam</servlet-name>
    <url-pattern>/MyServlet</url-pattern>
</servlet-mapping>

然后您可以使用MyServlet 作为表单的操作参数,以便按照上述servlet 映射将带有表单参数的请求发送给您的servlet SerExam

<form action="MyServlet" method="post">

当您在表单中使用POST 方法时,您的servlet doPost 方法将被调用,我可以从您的代码中看到您没有在doPost 方法中编写任何代码。

删除 doPost(request,response);(这是您的 doGet 方法中的第一行。

因此,将您的 servlet 代码更改为:

protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException

       //your existing code inside doGet with that doPost call removed   


protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException

     doGet(request,response);

在您的代码正常运行之前,您还需要完成许多事情。

所以我请你看看我们的Servlet wiki,这样你就可以得到基本的了解。

【讨论】:

【参考方案2】:

您的 servlet 映射到 /home.html:

<servlet-mapping>
    <servlet-name>SerExam</servlet-name>
    <url-pattern>/home.html</url-pattern>
</servlet-mapping>

当您访问firstserv时:

<form action="./firstserv" method="post">

旁注:

永远不要在您的数据库中以纯文本格式存储密码

ps=con.prepareStatement("select * from loginvalidation where username=? and password=?");
ps.setString(1, username);
ps.setString(2, password);

【讨论】:

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