Right now I have this code to open an image from an MDIParent Window called MDIParent1
private void OpenFile(object sender, EventArgs e)
{
OpenFileDialog openFileDialog = new OpenFileDialog();
openFileDialog.InitialDirectory = Environment.GetFolderPath(Environment.SpecialFolder.Personal);
openFileDialog.Filter = "Image Files (*.jpg)|*.jpg|All Files (*.*)|*.*";
if (openFileDialog.ShowDialog(this) == DialogResult.OK)
{
string FileName = openFileDialog.FileName;
Process.Start(@FileName);
}
}
This opens a new window with my image fine however I want this to open as a child window to MDIParent1. Any help would be greatly appreciated. Thank You
Process.Start(@FileName);
is the same like you double click the file from explorer. So it will open a new window. If you have set default program to open the image, then it will open the program instead.
If you want to do it by C#, create a form with a PictureBox in it. Then instead of calling Process.Start(@FileName);
, call the form like this:
Form1 form = new Form1();
form.MdiParent = this;
form.PictureBox1.ImageLocation = FileName;
form.Open();