Simple Windows Form Login Application in C#

Friday, January 16, 2015

Simple Windows Form Login Application in C#

By Anoop Kumar Sharma

C#, Windows Form Application

2 comments

In this Post, we will learn how to create a Simple Windows form Login application.

Let’s Begin:
1. Create a New Windows Form Application.

2. Add New Database (I have created a database named as MyDatabase.mdf). Add a table (named as tbl_Login). The following is the table schema for creating tbl_Login.

3. Create a form (frmLogin) and add Label, TextBox and button control from the Toolbox.

4. Add another Windows Form and named it as frmMain. This form will be shown to the user after successful Login by the user.

Now, Go to frmLogin.cs code and add System.Data and System.Data.SqlClient namespace. Double click on btn_Submit to create btn_Submit Click event.
frmLogin.cs Code:

using System; using System.Data; using System.Windows.Forms; using System.Data.SqlClient; namespace LoginApplication {     public partial class frmLogin : Form     {         public frmLogin()         {             InitializeComponent();         }         //Connection String         string cs = @"Data Source=(LocalDB)\v11.0;AttachDbFilename=|DataDirectory|\MyDatabase.mdf;Integrated Security=True;";         //btn_Submit Click event         private void button1_Click(object sender, EventArgs e)         {             if(txt_UserName.Text=="" || txt_Password.Text=="")             {                 MessageBox.Show("Please provide UserName and Password");                 return;             }             try             {                 //Create SqlConnection                 SqlConnection con = new SqlConnection(cs);                 SqlCommand cmd = new SqlCommand("Select * from tbl_Login where UserName=@username and Password=@password",con);                 cmd.Parameters.AddWithValue("@username",txt_UserName.Text);                 cmd.Parameters.AddWithValue("@password", txt_Password.Text);                 con.Open();                 SqlDataAdapter adapt = new SqlDataAdapter(cmd);                 DataSet ds = new DataSet();                 adapt.Fill(ds);                 con.Close();                 int count = ds.Tables[0].Rows.Count;                 //If count is equal to 1, than show frmMain form                 if (count == 1)                 {                     MessageBox.Show("Login Successful!");                     this.Hide();                     frmMain fm = new frmMain();                     fm.Show();                 }                 else                 {                     MessageBox.Show("Login Failed!");                 }             }             catch(Exception ex)             {                 MessageBox.Show(ex.Message);             }         }     } }

After that, go to frmMain.cs form and create btn_LogOut click event. On clicking the Logout button, frmMain form hides and show frmLogin form.

frmMain.cs Code:

using System; using System.Windows.Forms; namespace LoginApplication {     public partial class frmMain : Form     {         public frmMain()         {             InitializeComponent();         }         //btn_LogOut Click Event         private void btn_LogOut_Click(object sender, EventArgs e)         {             this.Hide();             frmLogin fl = new frmLogin();             fl.Show();         }         private void frmMain_FormClosing(object sender, FormClosingEventArgs e)         {             Application.Exit();         }     } }

Final Preview:

Hope you like it. Thanks.

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