C#打开文件所在的文件夹

              string filePath = dgvRecord.SelectedRows[0].Cells[3].Value.ToString();
                Process open = new Process();
                open.StartInfo.FileName = "explorer";
                open.StartInfo.Arguments = @"/select," + filePath;
                open.Start();

 

转载 http://www.cnblogs.com/illusion/archive/2011/04/16/2017890.html