how to open a windows form after a login form

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how to open a application which is a form after a login form.i have made a login form which results in a message login successfull in a messagebox.
can i open my application after that form when the login is successfull. and returns to the same login form if the login is not valid
my code for login form is
also my database is sql server 2008 and i am using visual studio
thanks in advance
help me plz

C#

private void button1_Click(object sender, EventArgs e)
        {
            SqlConnection con = new SqlConnection();
            con.ConnectionString = @"Data Source=.\SQLEXPRESS;AttachDbFilename=c:\documents and settings\aquib\my documents\visual studio 2010\Projects\login\login\Database1.mdf;Integrated Security=True;User Instance=True";
            try
            {
                con.Open();
            }
            catch (Exception)
            {
                MessageBox.Show("Error with the databse connection");
            }
            string qry1 = "Select * from Log007 where Password=@Password and Username=@Username";
            SqlCommand com = new SqlCommand(qry1, con);
            com.Parameters.AddWithValue("@Username", this.textBox1.Text);
            com.Parameters.AddWithValue("@Password", this.textBox2.Text);
            SqlDataReader dr = com.ExecuteReader();
            while (dr.Read())
            {
                if (dr.HasRows == true)
                {
                    MessageBox.Show("Login Successfull", "Login Information");
                }
            }
            if (dr.HasRows == false)
            {
                MessageBox.Show("Access Denied", "Login Information");
                this.Close();

Posted 18-Aug-12 18:01pm

shaikh-adil

Updated 18-Aug-12 18:25pm

Sergey Alexandrovich Kryukov

v2

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Comments

ridoy 19-Aug-12 2:52am

Simple solution...follow solution 2 posted here

2 solutions

Solution 3

use your code at Program class. In this code I assume that Form1 is your Start up WinForm.

C#

class Program
{
   int count = 3;
   do
   {
       // show log in form here 3 times if user get mistake at typing
       if (!dr.HasRows)
           Application.Exit();
       else
           break;

   } while (count != 0);

   Application.Run(new Form1());
}

Posted 18-Aug-12 20:48pm

WebMaster

Comments

shaikh-adil 19-Aug-12 2:55am

i have to write the whole button click code in program.cs?

[no name] 19-Aug-12 7:55am

I said "Show the login dialog at Program", not whole the project.

shaikh-adil 19-Aug-12 8:41am

bro i understood.
But my login dialog contain database connectivity.. Will it work?
I have my code above. Tell me it will work or not. And directly i have to write the whole code in program.cs file?

[no name] 19-Aug-12 13:59pm

Sure. I used that before myself.
There is no different where you write that.

shaikh-adil 19-Aug-12 22:50pm

okay i will give you feedback...
Will try after eid
thanx

shaikh-adil 19-Aug-12 23:02pm

bro this is my code in program.cs file

using System;
using System.Collections.Generic;
using System.Linq;
using System.Windows.Forms;
using System.Data.SqlClient;

namespace login
{
static class Program
{
///

/// The main entry point for the application.
///

[STAThread]
static void Main()
{
Application.EnableVisualStyles();
Application.SetCompatibleTextRenderingDefault(false);
Application.Run(new Form1());
}
private void button1_Click(object sender, EventArgs e)
{
SqlConnection con = new SqlConnection();
con.ConnectionString = @"Data Source=.\SQLEXPRESS;AttachDbFilename=c:\documents and settings\aquib\my documents\visual studio 2010\Projects\login\login\Database1.mdf;Integrated Security=True;User Instance=True";
try
{
con.Open();
}
catch (Exception)
{
MessageBox.Show("Error with the databse connection");
}
string qry1 = "Select * from Log007 where Password=@Password and Username=@Username";
SqlCommand com = new SqlCommand(qry1, con);
com.Parameters.AddWithValue("@Username", this.textBox1.Text);
com.Parameters.AddWithValue("@Password", this.textBox2.Text);
SqlDataReader dr = com.ExecuteReader();
while (dr.Read())
{
if (dr.HasRows == true)
{
MessageBox.Show("Login Successfull", "Login Information");
}
}
if (dr.HasRows == false)
{
MessageBox.Show("Access Denied", "Login Information");
this.Close();
}
}
}
}
many errors are commming,
i am correct??

[no name] 19-Aug-12 23:59pm

You didn't understand what I say.
compare your code with mine, did i say type your code at any button_Click?
You are totally wrong.

static void Main()
{
SqlConnection con = new SqlConnection();
//...
//...
while (dr.Read())
{
if (dr.HasRows == true)
{
MessageBox.Show("Login Successfull", "Login Information");
break;
}
}
if (dr.HasRows == false)
{
MessageBox.Show("Access Denied", "Login Information");
Application.Exit();
}

Application.EnableVisualStyles();
Application.SetCompatibleTextRenderingDefault(false);
Application.Run(new Form1());
}

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_Amy · Accepted Answer · 2012-08-18T18:26:00

Solution 2

Hi,
Try adding this when login is successful:

C#

if (dr.HasRows == true)
{
       MessageBox.Show("Login Successfull", "Login Information");
       this.Hide(); //This will hide your current form
       MyWinForm1 M1 = new MyWinForm1();//Create the object of your second form
       M1.Show(); //Show the second form using the object.
}

While closing the file handle the form closing event as I said in my comments below. Use this:

C#

private void Form1_Closing(object sender, System.ComponentModel.CancelEventArgs e)
{
  // Display a MsgBox asking the user to save changes or abort.
  if(MessageBox.Show("Do you want to save changes to your text?", "My Application",
     MessageBoxButtons.YesNo) ==  DialogResult.Yes)
  {
     // Cancel the Closing event from closing the form.
     e.Cancel = true;
     Application.Exit();
     // Call method to save file...
  }
}

All the best.
--Amit

Posted 18-Aug-12 18:26pm

_Amy

Updated 18-Aug-12 23:12pm

v2

Comments

[no name] 19-Aug-12 2:51am

Good Job.
My 5!

_Amy 19-Aug-12 3:02am

Thanks. :)

shaikh-adil 19-Aug-12 3:26am

hey when i closes the second form but the program is still running.i persnally hae to stop debugging why so bro

_Amy 19-Aug-12 3:36am

Use Application.Exit(); to exit the application. :)

shaikh-adil 19-Aug-12 3:58am

i know that but if i want to close the application by clicking the exit button which is there already there in the frame so how can i do that

_Amy 19-Aug-12 4:08am

Handle the Closing event of the form. In that use Application.Exit(); .

shaikh-adil 19-Aug-12 4:34am

you have not understood my question bro
i want to close the application by clicking X button bythe side of minimize and maximize.
i dont want to create a new button for closing.
help me in this

_Amy 19-Aug-12 5:12am

I understood your question. As I told handle the form closing event. Use my updated answer. :)

shaikh-adil 19-Aug-12 7:26am

bro there is a problem in that code.
I had handled what you told but. When i click the closing button a dialog appears which you have given that do you want to save changes to your text. That dialog appears and when i click yes then same dialog appears once more and when i click the same yess button it goes and my form is not going it is there in the visible. But when the dialog come and i clicked on no. Then the application exits but it is hiding.. I have to stop debugging every time.
Can u explain whats the problem

shaikh-adil 19-Aug-12 8:48am

bro this code works. Not on closing event but on closed event.
And i have commented e.cancle=true,
then the code works.
private void Form1_Closing( object sender, System.ComponentModel.CancelEventArgs e){ // Display a MsgBox asking the user to save changes or abort. if (MessageBox.Show( " Do you want to save changes to your text?" , " My Application" , MessageBoxButtons.YesNo) == DialogResult.Yes) { Application.Exit();
}