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Obtaining only the filename when using OpenFileDialog property “FileName”
3 answers
I have a save file dialog and i want to get only the filename entered. Equivalent for
openfiledialog.SafeFileName;
Save file dialog has no SafeFileName
Property and FileName
returns both filename, path and extension. Pls how do i extract only file name.
If you want the filename with extension use Path.GetFileName()
. If you want it without the extension as well use Path.GetFileNameWithoutExtension()
.
public void Test(string fileName)
{
string path = Path.GetDirectoryName(fileName);
string filename_with_ext = Path.GetFileName(fileName);
string filename_without_ext = Path.GetFileNameWithoutExtension(fileName);
string ext_only = Path.GetExtension(fileName);
}
See MSDN for further details, especially the Path
class which has a number of useful methods:
http://msdn.microsoft.com/en-us/library/System.IO.Path_methods.aspx
http://msdn.microsoft.com/en-us/library/system.io.path.getfilename.aspx
http://msdn.microsoft.com/en-us/library/system.io.path.getfilenamewithoutextension.aspx
Also found another solution to my problem
FileInfo fi = new FileInfo(saveFileDialog1.FileName);
string text = fi.Name;