如何更改“對象發送者”？

Question

我在makeDataGridView函數中創建了DataGridView，也許datagridview右鍵單擊->出現上下文菜單

這是示例代碼

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
        */
}

我想從'context_menu_click'函數中的datagridview打印一個.txt文件。 click函數的發送者是DataGridView，但是context_menu_click發送者是ToolStripMenu。

那么，如何在context_menu_click函數中從DataGridView打印.txt文件？

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)));
    }
}

您可以將當前的DataGridView作為Tag添加到剛創建的ToolStripItem

Answer 2

使用它來獲取點擊事件中的DataGridView

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