我需要获取启动事件的控件的名称,特别是我为DataGrid使用了ContextMenu,这是我实际编写的代码:
private void ClearTable_Click(object sender, RoutedEventArgs e)
{
// Try to cast the sender to a MenuItem
ContextMenu menuItem = sender as ContextMenu;
if (menuItem != null)
{
// Retrieve the ContextMenu that contains this MenuItem
ContextMenu menu = menuItem.GetContextMenu();
// Get the control that is displaying this context menu
Control sourceControl = menu.SourceControl;
}
}
XAML
<ContextMenu x:Key="Squadre_ContextMenu">
<MenuItem Header="Pulisci Tabella" Click="ClearTable_Click">
</MenuItem>
<MenuItem Header="Elimina Riga selezionata" Click="ClearRow_Click">
</MenuItem>
</ContextMenu>
但ContextMenu没有GetContextMenu方法,所以这对我来说是一个问题。有办法解决这个或其他方式吗?
答案 0 :(得分:2)
试试这个:
private void ClearTable_Click(object sender, RoutedEventArgs e)
{
// Try to cast the sender to a Control
Control ctrl = sender as Control;
if (ctrl != null)
{
// Get the control name
string name = ctrl.Name;
// Get parent control name
Control parent = (Control) ctrl.Parent;
string parentName = parent.Name
}
}
编辑:添加代码以获取父控件的名称
答案 1 :(得分:0)
将sender对象作为MenuItem然后获取TextBlock控件的ContextMenu的示例:
MenuItem mi = sender as MenuItem;
if (mi != null)
{
ContextMenu cm = mi.CommandParameter as ContextMenu;
if (cm != null)
{
TextBlock t = cm.PlacementTarget as TextBlock;
if (t != null)
{
// print t.Name or whatever...
}
}
}
希望这有帮助
答案 2 :(得分:0)
我创建了一个带有datagridview的表单,我得到了没有问题的名称
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;
namespace WindowsFormsApplication1
{
public partial class Form1 : Form
{
public Form1()
{
int l_Index = 0;
InitializeComponent();
MenuItem[] l_MenuItems = new MenuItem[3];
for (l_Index = 0; l_Index < l_MenuItems.Length; l_Index++)
{
l_MenuItems[l_Index] = new MenuItem("Menu " + l_Index.ToString(), MenuItem_Click);
}
System.Windows.Forms.ContextMenu l_ContextMenu = new ContextMenu(l_MenuItems);
dataGridView1.ContextMenu = l_ContextMenu;
}
private void MenuItem_Click(object sender, EventArgs e)
{
MenuItem l_MenuItem = sender as MenuItem;
if (l_MenuItem != null)
{
System.Windows.Forms.ContextMenu l_ContextMenu = l_MenuItem.GetContextMenu();
if (l_ContextMenu != null)
{
if (l_ContextMenu.SourceControl != null)
{
System.Diagnostics.Debug.WriteLine(l_ContextMenu.SourceControl.Name);
}
}
}
}
private void dataGridView1_MouseClick(object sender, MouseEventArgs e)
{
if (e.Button == MouseButtons.Right)
{
dataGridView1.ContextMenu.Show(dataGridView1, new Point(e.X, e.Y));
}
}
}
}