c# - OpenFileDialog exeでフォルダを開く

Question

OpenFileDialog を使用してアプリケーションのフォルダをすばやく開くにはどうすればよいですか?

        OpenFileDialog openFileDialog1 = new OpenFileDialog();
        openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
        if (openFileDialog1.ShowDialog() == DialogResult.OK)
        {
           ...........

        }

Answer 1

Guessing that you mean "Show the OpenFileDialog starting in my application's folder", just set the OpenFileDialog.InitialDir to your application's folder before showing the OpenFileDialog.

string AppPath = Path.GetDirectoryName(Application.ExecutablePath);;
openFileDialog1.InitialDir = AppPath;

If you need help finding your application's directory, see Getting root folder of application

Answer 2

Use the FileDialog.InitialDirectory Property if you wish to override one of the default ways for its value getting set (described on MSDN).

openFileDialog1.InitialDirectory = @"C:\";  // based on comment of question