【发布时间】:2016-01-01 09:34:05
【问题描述】:
我对 Web 开发完全陌生,我需要一些帮助。我正在使用 Java Eclipse EE、tomcat 服务器和 mysql 做一个工资系统 web 应用程序项目。我使用了一个教程并设法在下面创建了登录界面。所以现在,当我单击输入我的登录详细信息并单击登录(在 localhost:8080/Payroll)时,我希望它转到一个网页(我不知道如何创建)并显示按钮列表(任何随机我以后可以重命名的按钮)。有人可以帮帮我吗。我不知道如何使用 .JSP、.html、.java,而且我对这些文件类型如何帮助我获得我想要的东西感到非常困惑。请帮助某人,我只想将登录按钮重定向到带有按钮的网页。谢谢。
Login.java (Servlet)
import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;
import java.sql.*;
public class Login extends HttpServlet {
protected void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
response.setContentType("text/html;charset=UTF-8");
PrintWriter out = response.getWriter();
String employee_id = request.getParameter("employee_id");
String password = request.getParameter("password");
if(Validate.checkUser(employee_id, password)) {
RequestDispatcher rs = request.getRequestDispatcher("**SOME FILE NAME HERE TO REDIRECT TO?**");
rs.forward(request, response);
}
else
{
out.println("Employee ID or Password is incorrect. Please try again.");
RequestDispatcher rs = request.getRequestDispatcher("index.html");
rs.include(request, response);
}
}
}
index.html
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<title>Login</title>
</head>
<body>
<form action="login" method="post">
<h3>
Employee Login
</h3>
<b>Employee ID:</b> <br>
<input type="text"name="employee_id" size="20"><br><br>
<b>Password:</b><br>
<input type="password" name="password" size="20"><br><br>
<input type="submit" value="Login"><br><br>
</form>
</body>
</html>
Validate.java(类文件)
import java.sql.*;
public class Validate
{
public static boolean checkUser(String employee_id, String password)
{
boolean st = false;
try {
Class.forName("com.mysql.jdbc.Driver").newInstance();
Connection con = DriverManager.getConnection("jdbc:mysql://localhost:3306/payroll_system", "root", "");
PreparedStatement ps = con.prepareStatement("select * from employee_login where employeeID = ? and pwd = ?");
ps.setString(1, employee_id);
ps.setString(2, password);
ResultSet rs =ps.executeQuery();
st = rs.next();
}catch(Exception e)
{
e.printStackTrace();
}
return st;
}
}
【问题讨论】:
标签: java mysql jsp tomcat servlets