Also HTTP Status 404 - /Login/userdetail.java

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ikramlim

Also HTTP Status 404 - /Login/userdetail.java
1 Answer(s) 5 years and 2 months ago
Posted in : JSP-Servlet

same can't to run.. error also please help..

index.jsp

<%@page pageEncoding="UTF-8"%>
<html>
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
        <title>JSP Page</title>
    </head>
    <body>
        <form name="f1" action="userdetail.java" method="post">
            Username:<input type="text" name="uname">
            Password:<input type="text" name="upass">
                     <input type="submit" value="Send">
            </form>
    </body>
</html>

userdetail.java

(server I'm using wamp server to connect. add library - MySQL JDBC Driver-mysql-connector-java-5.1.13-bin.jar - apache tomcat

In the services, in NetBeans IDE already at server -jdbc:mysql://localhost:3306/userdetail[root on Default schema

 
import java.io.*; 
import java.net.*; 
import javax.servlet.*; 
import javax.servlet.http.*; 
import java.sql.*;

@WebServlet(name = "userdetail", urlPatterns = {"/userdetail"}) public class userdetail extends HttpServlet {

 public class LoginServlet extends HttpServlet{
      public void doPost(HttpServletRequest request,
      HttpServletResponse response)
      throws ServletException, IOException {
      response.setContentType("text/html");
      PrintWriter out = response.getWriter();
      String user=request.getParameter("uname");
      String pass=request.getParameter("upass");

   try {

   Class.forName("com.mysql.jdbc.Driver").newInstance();
            Connection myConn=DriverManager.getConnection("jdbc.mysql://localhost:3306/Login","root","root");
            Statement myStmt=myConn.createStatement();
            String query="select * from user where username='"+user+"' and password='"+pass+"'";
            ResultSet myRs=myStmt.executeQuery(query);

          if(myRs.next());
        {
            out.println("Welcome "+user);
        }

        }
    catch(Exception e)
    {
        System.out.println("Exception"+e);
    }


  }
}

*please help.. Thx alot

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July 24, 2012 at 11:16 AM

Post your error that comes while running the application.

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