Can't set the position of a contextmenustrip?

Question

I'm trying to open a contextmenustrip at the place where I right-clicked the mouse, but it always shows at top left of the screen.

Here is the code I used:

private void dataGridView1_RowHeaderMouseClick(object sender, DataGridViewCellMouseEventArgs e)
{
    if (e.Button == MouseButtons.Right)
    {
        contextMenuStrip1.Show(new Point(e.X,e.Y));
        doss.getdossier(connection.conn, Convert.ToInt32(dataGridView1.Rows[e.RowIndex].Cells[0].Value));
    }
}

your platform would help alot. (silverlight,wpf,asp.net,winforms, etc) — Muad'Dib, Sep 13 '11 at 15:26

score 12 · Accepted Answer · answered Sep 13 '11 at 15:30

if (e.Button == MouseButtons.Right)
{
    contextMenuStrip1.Show(Cursor.Position);
}

the reason it's not appearing is because you are using e.X and e.Y for the values. They are not the actual location on the screen. They are the location of the mouse within the datagrid. So say you clicked on the first cell of the first row, that will be near the top left of that component. e.X and e.Y are the mouse locations within the component.

score 2 · Answer 2 · answered Sep 13 '11 at 15:32

assuming you are in Windows Forms, try this:

if (e.Button == MouseButtons.Right)
{
  Control control = (Control) sender;

  // Calculate the startPoint by using the PointToScreen 
  // method.

  var startPoint = control.PointToScreen(new Point(e.X, e.Y));
  contextMenuStrip1.Show(startPoint);
  ...
  ...

Can't set the position of a contextmenustrip?

2 Answers2