我设计了一个包含面板和标签的表单。 我手工制作了面板的光标属性。 但是当我点击面板时,我得到了这个例外:
无法转换类型为#System; Windows.Forms.Panel'的对象。输入' System.Windows.Forms.Label'
这是我的代码:
using System;
using System.Drawing;
using System.Windows.Forms;
namespace Mancula
{
public partial class Form2 : Form
{
private System.Windows.Forms.Panel[] pan;
private System.Windows.Forms.Label[] lab;
Graphics g;
Pen pen = new Pen(Color.Red);
SolidBrush br = new SolidBrush(Color.Red);
Rectangle rec = new Rectangle(10, 10, 30, 30);
int player1 = 49;
int player2 = 49;
int playerturn = 0;
public Form2()
{
InitializeComponent();
label1.Text = "0";
label2.Text = "0";
}
public void startgame()
{
if (radioButton1.Checked == true)
{
playerturn = 1;
}
else
{
playerturn = 2;
}
}
private void button15_Click(object sender, EventArgs e)
{
button1.Enabled = button2.Enabled = true;
button3.Enabled = button4.Enabled = true;
button5.Enabled = button6.Enabled = true;
button7.Enabled = button8.Enabled = true;
button9.Enabled = button10.Enabled = true;
button11.Enabled = button12.Enabled = true;
button13.Enabled = button14.Enabled = true;
panel1.Enabled = panel2.Enabled = true;
panel3.Enabled = panel4.Enabled = true;
panel5.Enabled = panel6.Enabled = true;
panel7.Enabled = panel8.Enabled = true;
panel9.Enabled = panel10.Enabled = true;
panel11.Enabled = panel12.Enabled = true;
panel13.Enabled = panel14.Enabled = true;
label1.Text = player1.ToString();
label2.Text = player2.ToString();
startgame();
panel15.Enabled = false;
for (int i = 0; i < 14; i++)
{
pan [i].Click += new System.EventHandler(this.ClickHandler);
}
}
private void ClickHandler(object sender, EventArgs e)
{
Panel temppanel = (Panel)sender;
Label templab = (Label)sender;
if (playerturn == 1)
{
if (player1 == 0)
{
playerturn = 2;
}
else
{
player1 = player1 - 1;
if (templab.Text != "7")
{
g = temppanel.CreateGraphics();
g.DrawRectangle(pen, rec);
label1.Text = player1.ToString();
}
}
}
else
{
if (player2 == 0)
{
playerturn = 1;
}
else
{
player2 = player2 - 1;
if (templab.Text != "7")
{
g = temppanel.CreateGraphics();
g.DrawRectangle(pen, rec);
label2.Text = player2.ToString();
}
}
}
}
private void Form2_Load(object sender, EventArgs e)
{
pan = new Panel[14] { panel1, panel2, panel3, panel4, panel5, panel6,panel7,
panel8,panel9,panel10,panel11,panel12,panel13,panel14
};
lab = new Label[14] { lab1,lab2,lab3,lab4,lab5,lab6,lab7,lab8,lab9,lab10,lab11,lab12,lab13,lab14};
}
}
}
答案 0 :(得分:2)
我认为你的错误是这一行:
Panel temppanel = (Panel)sender;
Label templab = (Label)sender;
你应该在进行演员表之前检查一下类型。
你可以这样做:
if(sender.GetType() == typeof(Panel))
{
Panel temppanel = (Panel)sender;
}
else
{
Label templab = (Label)sender;
}
无论如何,将相同的处理程序分配给不同类型的对象似乎不是一个好主意。我的建议是为每种类型的控件创建一个处理程序。
答案 1 :(得分:1)
在您的事件处理程序中,您有:
Panel temppanel = (Panel)sender;
Label templab = (Label)sender;
但发件人可以是Panel
或Label
,这就是为什么其中一个演员会失败的原因。
答案 2 :(得分:0)
你的问题就在这一行:
Label templab = (Label)sender;
当您点击该面板时,sender
为Panel
,而不是Label
,因此您无法将其投射。