I'm using c# to make a mobile 6 application. I created another windows form in the project. This is the form that I would like to load first. This is what have tried:
MainMenu gameMenu = new MainMenu();
private void Main开发者_高级运维Form_Load(object sender, EventArgs e)
{
this.Hide();
gameMenu.ShowDialog();
.....
}
When I run this the emulator comes up but it just stays as the default windows screen. And I don't get any of my forms.
GameMenu's parent is MainForm, which is now hidden, so the Dialog isn't going to be visible. You need to adjust your logic to do one of the following:
- show the GameMenu first (i.e.
Application.Run(new GameMenu)
) - Don't hide MainForm
- Use
gameMenu.Show()
instead ofShowDialog()
You may need to get rid of this.Hide() or use gameMenu.Show() instead of gameMenu.ShowDialog() or you may need to do both.
If you have to use gameMenu.Show() instead of gameMenu.ShowDialog(), you may also want to do the following: Subscribe to MainForm's GotFocus event and call gameMenu.Show() again whenever the other form gains focus unintentionally. Set MainForm's Enabled property to false while the gameMenu is shown if you want to prevent any accidental interaction with the MainForm while the gameMenu is supposed to be shown.
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