Extracting Path from OpenFileDialog path/filename


I'm writing a little utility that starts with selecting a file, and then I need to select a folder. I'd like to default the folder to where the selected file was.

OpenFileDialog.FileName returns the full path & filename - what I want is to obtain just the path portion (sans filename), so I can use that as the initial selected folder.

    private System.Windows.Forms.OpenFileDialog ofd;
    private System.Windows.Forms.FolderBrowserDialog fbd;
    if (ofd.ShowDialog() == DialogResult.OK)
        string sourceFile = ofd.FileName;
        string sourceFolder = ???;
    fbd.SelectedPath = sourceFolder; // set initial fbd.ShowDialog() folder
    if (fbd.ShowDialog() == DialogResult.OK)

Are there any .NET methods to do this, or do I need to use regex, split, trim, etc??

2/13/2018 9:50:24 AM

Accepted Answer

Use the Path class from System.IO. It contains useful calls for manipulating file paths, including GetDirectoryName which does what you want, returning the directory portion of the file path.

Usage is simple.

string directoryPath = Path.GetDirectoryName(filePath);
1/13/2009 2:03:06 PM

how about this:

string fullPath = ofd.FileName;
string fileName = ofd.SafeFileName;
string path = fullPath.Replace(fileName, "");

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