How to change the 'object sender'?

Question

I create DataGridView in makeDataGridView function, and maybe datagridview rightclick -> appear context menu

This is sample code

public void click(object sender, MouseEventArgs e) {
        if(e.button == MouseButtons.Right) {
               ContextMenuStrip menu = new ContextMenuStrip();
               ToolStripItem insert = menu.Items.Add("insert");
               insert.Click += new EventHandler(context_menu_click);
        }
}

public void context_menu_click(object sender, EventArgs e) {
        / *
           Other event and
           printing .txt file from DataGridView
        */
}

I want to print a .txt file from datagridview in 'context_menu_click' function. The sender of click function is DataGridView, but context_menu_click sender is ToolStripMenu.

So, How can I print a .txt file from DataGridView in context_menu_click function?

Answer 1

public void context_menu_click(object sender, EventArgs e)
{
    /*
        Other event and
        printing .txt file from DataGridView
    */
    var item = sender as ToolStripItem  ;
    if (item != null)
    {
        DataGridView gv = item.Tag as DataGridView;
        Console.WriteLine(gv.Name);
    }
}
private void dataGridView1_MouseClick(object sender, MouseEventArgs e)
{
    if (e.Button == MouseButtons.Right)
    {
        ContextMenuStrip menu = new ContextMenuStrip();
        ToolStripItem insert = menu.Items.Add("insert");
        insert.Tag = sender;
        insert.Click += new EventHandler(context_menu_click);
        menu.Show(this.dataGridView1.PointToScreen(new Point(e.X, e.Y)));
    }
}

You can add current DataGridView as Tag to the ToolStripItem you just create

Answer 2

使用它来获取点击事件中的DataGridView

((ContextMenuStrip)((ToolStripMenuItem)sender).GetCurrentParent()).SourceControl