如何将事件处理程序作为方法参数传递?

时间:2010-02-01 11:14:16

标签: c# parameters delegates methods

如何将事件处理程序 TextBlock_MouseDown_Test1 TextBlock_MouseDown_Test2 传递给SmartGrid,以便它创建的TextBlocks在单击时会执行此事件处理程序?

以下代码收到错误:

  

最佳重载方法匹配   “TestDel234.SmartGrid.SmartGrid(TestDel234.Window1,   System.Collections.Generic.List,   System.EventHandler)'有一些无效   参数

using System;
using System.Collections.Generic;
using System.Windows;
using System.Windows.Controls;
using System.Windows.Documents;
using System.Windows.Input;

namespace TestDel234
{
    public partial class Window1 : Window
    {
        public Window1()
        {
            InitializeComponent();
            List<string> items = new List<string> { "one", "two", "three" };
            SmartGrid sg = new SmartGrid(this, items, TextBlock_MouseDown_Test1);
        }

        private void TextBlock_MouseDown_Test1(object sender, MouseButtonEventArgs e)
        {
            MessageBox.Show("testing1");
        }

        private void TextBlock_MouseDown_Test2(object sender, MouseButtonEventArgs e)
        {
            MessageBox.Show("testing2");
        }
    }

    public class SmartGrid
    {
        public SmartGrid(Window1 window, List<string> items, EventHandler eventHandler)
        {
            foreach (var item in items)
            {
                TextBlock tb = new TextBlock();
                tb.Text = item;
                tb.MouseDown += new MouseButtonEventHandler(eventHandler);
                window.MainContent.Children.Add(tb);
            }
        }
    }
}

1 个答案:

答案 0 :(得分:7)

您不能将鼠标按钮事件args处理程序转换为普通的EventHandler - 而是在构造函数中尝试EventHandler<MouseButtonEventArgs>