ShowDialog property for MDI winforms

Question

Question:

How do I display a MDI child form in a ShowDialog() format?

What I've tried:

private void Add()
        {

            ModuleAddPopUp map = new ModuleAddPopUp();
            map.StartPosition = FormStartPosition.CenterScreen;
            map.ShowDialog();          
        }

Doing the above, the form displays center screen as a pop-up, however I can drag the form outside the MDI when the MDI isn't maximized.

private void Add()
        {
            ModuleAddPopUp map = new ModuleAddPopUp();
            FormFunctions.OpenMdiDataForm(App.Program.GetMainMdiParent(), map);

        }

Doing the above, the form displays center screen, doesn't allow for the form to be dragged outside the MDI, but acts as a map.Show() , rather than a map.ShowDialog();

Answer 1

Add this code to your ModuleAddPopup class:

protected override void WndProc(ref Message message)
{
    const int WM_SYSCOMMAND = 0x0112;
    const int SC_MOVE = 0xF010;
    //SC_SIZE = 0XF000 if you also want to prevent them from resizing the form.
    //Add it to the 'if' condition.
    switch (message.Msg)
    {
        case WM_SYSCOMMAND:
            int command = message.WParam.ToInt32() & 0xfff0;
            if (command == SC_MOVE)
                return;
            break;
    }

    base.WndProc(ref message);
}

This is native code wrapped in C# code, as seen in here. This, however, will prevent the user from moving the dialog form anywhere.

Then, in your main form:

private void Add()
{
    ModuleAddPopUp map = new ModuleAddPopUp();
    map.StartPosition = FormStartPosition.CenterParent;
    map.ShowDialog();          
}