Extracting Path from OpenFileDialog path/filename

user_10410 0
Viewed 271980

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??

path parsing .net c#
6 Answers

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);
Jeff_232340

This was all I needed for the full path to a file

@openFileDialog1.FileName
user_26947200
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