OpenFileDialog1 C# OpenFileDialog C# OpenFileDialog OpenFileDialog dlg = new OpenFileDialog(); dlg.Multiselect = true; if (dlg.ShowDialog() == true) { var result = await Upload(new List(dlg.SafeFileNames), new List(dlg.FileNames)); MessageBox.Show(result ? "성공" : "실패"); } 2020. 7. 27. 이전 1 다음