Пример #1
0
        private void openToolStripMenuItem_Click(object sender, EventArgs e)
        {
            DialogResult dr = opn.ShowDialog();

            if (dr == System.Windows.Forms.DialogResult.OK)
            {
                string fname = opn.FileName;
                lf = new PCL.OutputLogging.LogFile(fname);
                fillDG();
            }
        }
Пример #2
0
 public void openfile(string fname)
 {
     lf = new PCL.OutputLogging.LogFile(fname);
     fillDG();
 }