OpenFileDialog open folder with exe

Question

How can I promptly open application's folder using OpenFileDialog ?

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

        }

i call OpenFileDialog and it shows me a folder with my exe, not c:\\ — user1872295, Dec 03 '12 at 23:14
Your question is still not clear. Do you mean "OpenFileDialog starts in my application's folder"? Please edit your question and add the information there, instead of adding it in the comments, so that people can see it. — Ken White, Dec 03 '12 at 23:18

score 6 · Answer 1 · edited May 23 '17 at 10:24

6

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

edited May 23 '17 at 10:24

Community

1
1

answered Dec 03 '12 at 23:27

Ken White

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score 2 · Answer 2 · answered Dec 03 '12 at 23:26

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

answered Dec 03 '12 at 23:26

Austin Salonen

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OpenFileDialog open folder with exe

2 Answers2