Login in console using jdbc

In this blog, we will learn how to validate a user name and password by using a table named login having columns as uname and pass using JDBC.

Introduction 

 

In this blog, we will know to validate user name and password by using a table named as login having columns as uname and pass using jdbc. If valid credentials are provided then it will show the welcome: username otherwise Invalid user name and password.

 

Here we use Type-1 driver (JDBC-ODBC bridge)

 

Creation of dsn(database source name) for Oracle

 

Start-Control panel- Administrative Tools- Data Sources (ODBC)-go to system dsn tab-click add button-select a driver for which you want to set up a data source (for Oracle- Oracle in XE)-select it and click finish-give any name in the data source name textbox-then click ok button.

 

Note: - Here Username=system, Password=pintu and Dsn name= dsnlogin

 

Table Creation with values

create table login(uname varchar(50), pass varchar(50))
insert into login values('raj', 'raj123')
insert into login values('ravi', 'ravi123')
insert into login values('rahul', 'rahul123')
import java.sql.*;
import java.util.*;
public class login {
 public static void main(String args[]) throws Exception
 {
  Class.forName("sun.jdbc.odbc.JdbcOdbcDriver");
  Connection con = DriverManager.getConnection("jdbc:odbc:dsnlogin", "system", "pintu");
  Statement stmt = con.createStatement();
  Scanner sc = new Scanner(System.in);
  System.out.print("Enter the user id : ");
  String str1 = sc.next();
  System.out.print("Enter the password : ");
  String str2 = sc.next();
  ResultSet rs = stmt.executeQuery("select * from login where uname='" + str1 + "' and pass='" + str2 + "'");
  if (rs.next())
   System.out.println("Welcome::: " + str1);
  else
   System.out.println("Invalid user name and password");
  con.close();
 }
}

Compile

 

Javac login.java

 

Java login