Single click to open menu for tray icon in C#
-
08-07-2019 - |
Question
How do i force a context menu for a tray icon to be shown when it is click rather than just right-clicked.
Ive tried using the MouseClick event, but the eventargs have the mouse position at x0y0.
Solution
This should do it for you:
private void notifyIcon1_Click(object sender, EventArgs e)
{
contextMenuStrip1.Show(Cursor.Position.X, Cursor.Position.Y);
}
OTHER TIPS
An alternate method that I have found to work a bit better:
private void notifyIcon1_MouseUp(object sender, MouseEventArgs e)
{
if (e.Button == MouseButtons.Left)
{
System.Reflection.MethodInfo mi = typeof(NotifyIcon).GetMethod("ShowContextMenu", System.Reflection.BindingFlags.Instance | System.Reflection.BindingFlags.NonPublic);
mi.Invoke(notifyIcon1, null);
}
}
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