C# Switching forms with passed data in the constructor

0
votes

I'm searching for the proper way how to open and switch forms some uses Application.Run( new Form1());, some uses Form1.ShowDialog(); and Form.Show();. I really want to know how to properly pass a data from a form into another using constructors.

Additionally I want to know why when I use Form.Close(); to close to current form before to open next form .... both forms closes.

Here are my codes.

            try
            {
                Form2 f2 = new Form2(connection, userLogin);
                this.Hide();
                f2.ShowDialog();
            }
            catch (NullReferenceException nre) {
                MessageBox.Show("Sorry Login Another User account");
            }

What I'm trying to do here is to pass the MySqlConnection in the variable connection and the valid user in the variable userLogin into the Form2. This method works but I'm not sure if this the right way to do it.

Here are the codes in Form2. 

public partial class Form2 : Form
    {
        MySqlConnection connection;
        User activeUser;

        public Form2(MySqlConnection pConn, User loginUser)
        {
            InitializeComponent();
            connection = pConn;
            activeUser = loginUser;
            this.Init();
            this.CenterToScreen();

        }


        private void logoutB_Click(object sender, EventArgs e)
        {
            this.Hide();
            new Form1().Visible = false;
            new Form1().ShowDialog();

            //Application.Run(new Form1());
        }
    }

Showing the Form2 doesn't also have a problem but when the logout button is pressed.

Form that is already visible cannot be displayed as a modal dialog box. 
Set the form's visible property to false before calling showDialog.

It says Form already visible? So does it mean the form is still open even though I used this.Hide();? and when I use the code Application.Run(new Form1());

Starting a second message loop on a single thread is not a valid operation. 
Use Form.ShowDialog instead.
2
c#formswinforms
Can you please explain the relationship between Form1 and Form2? Are you supposed to go back to the same Form1 (the same instance) when you close Form2? If so, you need to use ShowDialog when you show Form2. – Yacoub Massad
I will say that Form1 is the Login Screen and the Form2 as the Main Menu for the Application and Yes, I need to go back the same instance of Form1 when I click the Logout button on Form2. – ZeroCool

2 Answers

2
votes

In this section of code:

new Form1().Visible = false;
new Form1().ShowDialog();

You are simply creating two new forms; you are creating a form in the first line with Visibilityproperty set to false; and in the second line you are creating a new Form and calling ShowDialog on it. So two instances are not the same.

this.Hide();
Form1 a = new Form1();
a.Visible = false;
a.ShowDialog();
1
votes

Since you use ShowDialog to show Form2 from Form1, then simply close the second form when the user clicks the logout button.

private void logoutB_Click(object sender, EventArgs e)
{
    this.Close();
}

This will bring execution back to Form1 right after the ShowDialog call. So, make sure then you re-show the form after you invoke ShowDialog like this:

try
{
    Form2 f2 = new Form2(connection, userLogin);
    this.Hide(); //hide my self
    f2.ShowDialog(); //show Form2.

    //Execution will resume here after Form2 is closed
    this.Show(); //re-show my self
}
catch (NullReferenceException nre)
{
    MessageBox.Show("Sorry Login Another User account");
}