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??
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);
how about this:
string fullPath = ofd.FileName;
string fileName = ofd.SafeFileName;
string path = fullPath.Replace(fileName, "");
if (openFileDialog1.ShowDialog(this) == DialogResult.OK)
{
strfilename = openFileDialog1.InitialDirectory + openFileDialog1.FileName;
}