我在DataGridView
控件中使用了一个下拉列表,但问题是我第一次单击下拉列表时,需要两次单击才能下拉列表并显示,但之后它工作正常。
private void ViewActiveJobs_CellClick(object sender, DataGridViewCellEventArgs e)
{
if (e.RowIndex>=0)
{
jobCardId = int.Parse(ViewActiveJobs.Rows[ViewActiveJobs.CurrentCell.RowIndex].Cells["Job Card Number"].Value.ToString());
RegNo = ViewActiveJobs.Rows[ViewActiveJobs.CurrentCell.RowIndex].Cells["Registeration Number"].Value.ToString();
SelectedRow = e.RowIndex;
}
}
private void ViewActiveJobs_EditingControlShowing(object sender, DataGridViewEditingControlShowingEventArgs e)
{
try
{
ComboBox cbox = (ComboBox)e.Control;
cbox.SelectedIndexChanged -= new EventHandler(comboBOX_SelectedIndexChanged);
cbox.SelectedIndexChanged += new EventHandler(comboBOX_SelectedIndexChanged);
}
catch(Exception)
{
}
}
private void comboBOX_SelectedIndexChanged(object sender, EventArgs e)
{
ComboBox combo = sender as ComboBox;
string str = combo.SelectedIndex.ToString();
if (combo.SelectedIndex ==1)
pdf = new MakePDF(jobCardId,RegNo);
if (combo.SelectedIndex == 2)
{
PdfJobCard = new MakePDFJobCard(jobCardId);
}
if (combo.SelectedIndex == 3)
{
if (MessageBox.Show("Are you Sure you want to Close Job Card ?", "Are you Sure",
MessageBoxButtons.YesNo, MessageBoxIcon.Question) == DialogResult.Yes)
{
cmd = new SqlCommand();
cmd.Connection = con;
cmd.Parameters.Add("@jCard", SqlDbType.VarChar).Value = jobCardId;
cmd.Parameters.Add("@stat", SqlDbType.VarChar).Value = "Closed";
cmd.CommandText = "UPDATE JobCard SET status = @stat WHERE Id = @jCard";
try
{
cmd.ExecuteNonQuery();
ViewActiveJobs.Visible = false;
ViewActiveJobs.AllowUserToAddRows = true;
ViewActiveJobs.Rows.RemoveAt(SelectedRow);
//ViewActiveJobs.Visible = true;
}
catch (Exception c)
{
MessageBox.Show(c.Message);
}
}
}
}
答案 0 :(得分:4)
这是预期的行为。第一次点击是将焦点设置到组合框所必需的。一旦控件具有焦点,第二次单击就会显示下拉列表。
这会回答你的问题吗?或者您是否觉得有必要覆盖默认行为?在回答“是”之前,请考虑键盘用户以及使用箭头键在DataGridView
中从单元到单元格导航的用户。
如果答案仍然是肯定的,请参阅my answer此相关问题。基本上,您需要确保DataGridView
控件的EditMode
property设置为“EditOnEnter”,然后虚拟“按下”{{3中的 F4 键用于下拉组合框的事件处理程序。
暂且不说: 你的代码中应该 块为空Catch
块!解决了这个问题。