Пример #1
0
 private void editDataToolStripMenuItem_Click(object sender, EventArgs e)
 {
     if (editDataForm == null || editDataForm.IsDisposed)
     {
         editDataForm = new EditDataForm(this);
         editDataForm.Show();
     }
     else if (!editDataForm.IsDisposed)
     {
         editDataForm.Focus();
     }
 }
 private void ShowEditDataForm(EditDataForm.Tab startTab)
 {
     try
     {
         editDataForm = new EditDataForm(OpenedData);
         editDataForm.SetActiveTab(startTab);
         editDataForm.ShowDialog(this);
         if (editDataForm.DataChanged)
         {
             OpenedDataChanged = true;
             OpenedData = editDataForm.NewData;
         }
     }
     catch (Exception ex)
     {
         MessageBox.Show(ex.Message + ":\n" + ex.StackTrace, "Ошибка при работе с данными",
                         MessageBoxButtons.OK, MessageBoxIcon.Error);
     }
 }