Save data entered from two forms

Kishore Konatham

Hi everyone,

I've tried to save details in a data table from two winforms. It is saving form1 data in 1st row and form2 data in 2nd row. But i want it to save in a single row. Can you guys suggest me how?

//FORM1 CODE

private void button1_Click(object sender, EventArgs e)

{

string c = @"Data Source=mysql; Initial Catalog=SampleDB; Integrated Security=True";

Form2 f2 = new Form2();

SqlConnection con = new SqlConnection(c);

con.Open();

SqlCommand cmd = new SqlCommand("studentdetails", con);

cmd.CommandType = CommandType.StoredProcedure;

cmd.Parameters.AddWithValue("@Name",txtName.Text.Trim());

cmd.Parameters.AddWithValue("@Contact",txtContact.Text.Trim());

cmd.Parameters.AddWithValue("@Email", f2.txtEmail.Text.Trim());

cmd.Parameters.AddWithValue("@Address", f2.txtAddress.Text.Trim());

cmd.ExecuteNonQuery();

MessageBox.Show("Details Saved..!");

}

private void button2_Click(object sender, EventArgs e)

{

Form2 f2 = new Form2();

f2.Show();

}

//FORM2 CODE

private void button1_Click(object sender, EventArgs e)

{

string c = @"Data Source=mysql; Initial Catalog=SampleDB; Integrated Security=True";

Form1 f1=new Form1();

SqlConnection con = new SqlConnection(c);

con.Open();

SqlCommand cmd = new SqlCommand("studentdetails", con);

cmd.CommandType = CommandType.StoredProcedure;

cmd.Parameters.AddWithValue("@Name",f1.txtName.Text.Trim() );

cmd.Parameters.AddWithValue("@Contact",f1.txtContact.Text.Trim());

cmd.Parameters.AddWithValue("@Email", txtEmail.Text.Trim());

cmd.Parameters.AddWithValue("@Address", txtAddress.Text.Trim());

cmd.ExecuteNonQuery();

MessageBox.Show("Details Saved..!");

}

5 Replies

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Hemant Jindal7y ago
you can use singleton design pattern to have shared data across these two forms
0
Priyanka Jain7y ago
1. //FORM1 CODE
2. public static string Name = "";
3. public static string Contact = "";
5. private void button1_Click(object sender, EventArgs e)
6. {
7. string c = @"Data Source=mysql; Initial Catalog=SampleDB; Integrated Security=True";
8. Form2 f2 = new Form2();
9. Name = txtName.Text.Trim();
10. Contact = txtContact.Text.Trim();
11. MessageBox.Show("Details Saved..!");
12. }
14. private void button2_Click(object sender, EventArgs e)
15. {
16. Form2 f2 = new Form2();
17. f2.Show();
18. }
19. //FORM2 CODE
20. private void button1_Click(object sender, EventArgs e)
21. {
22. string c = @"Data Source=mysql; Initial Catalog=SampleDB; Integrated Security=True";
23. Form1 f1=new Form1();
24. SqlConnection con = new SqlConnection(c);
25. con.Open();
26. SqlCommand cmd = new SqlCommand("studentdetails", con);
27. cmd.CommandType = CommandType.StoredProcedure;
28. cmd.Parameters.AddWithValue("@Name",f1.Name );
29. cmd.Parameters.AddWithValue("@Contact",f1.Contact);
30. cmd.Parameters.AddWithValue("@Email", txtEmail.Text.Trim());
31. cmd.Parameters.AddWithValue("@Address", txtAddress.Text.Trim());
32. cmd.ExecuteNonQuery();
33. MessageBox.Show("Details Saved..!");
34. }
Try above code it may resolve your issue
0

Bhuban Magar7y ago

Hi,

Please return Current Insured ID from Stored procedure.

DECLARE @ID INT;

SET @ID=SCOPE_IDENTITY();

SELECT @ID;

Then,

On form1:

First Declare variable,

Private ID INT=0;

Then on Button Click Event:

int ID=cmd.ExecuteNonQuery();

con.close();

Pass Return ID as parameter:

private void button2_Click(object sender, EventArgs e)

{

Form2 f2 = new Form2(ID);

f2.Show();

}

On Form2:

write this on Form2 Constructor:

Private int _id=0;

Public Form2(int ID)

{

InitializeComponent();

this._id=ID;

}

Then Pass ID parameter also

cmd.Parameters.AddWithValue("@ID",_id);

update that table with this ID.

Hope this code will help you.

Thanks.

Kishore Konatham7y ago

Every time i can not update each row, if i use ID.

string qry = "update details set Email='" + f2.txtEmail.Text.Trim() + "', Address='" + f2.txtAddress.Text.Trim() + "' where Id=Id"; ???

Amit Kumar Singh7y ago

Hi Kishore,

Take Id from row which is inserting first and on basis of that u can update second form that in that row.

If I am wrong,Correct me !!!

Regards,

Amit Kumar Singh

+91-8291406438