Question

很抱歉再次打扰一个类似的问题。但是，我在C＃中再次遇到我的代码问题。我想告诉你到目前为止的代码：

    private void pictureBox2_Click(object sender, EventArgs e)
    {
        PictureBox flower2 = new PictureBox();
        flower2.Image = Properties.Resources.Flower3;
        flower2.Location = new Point(panel1.Location.X + 10, panel1.Location.Y + 10);
        flower2.Size = new System.Drawing.Size(pictureBox2.Size.Width, pictureBox2.Size.Height);
        flower2.Parent = panel1;
        this.Controls.Add(flower2);

        flower2.BringToFront();
        flower2.MouseMove += new MouseEventHandler(flower2_MouseMove);
        flower2.MouseDown += new MouseEventHandler(flower2_MouseDown);
    }

    private void flower2_MouseDown(object sender, MouseEventArgs e)
    {
        if (e.Button == System.Windows.Forms.MouseButtons.Left)
        {
            MouseDownLocation = e.Location;
        }
    }

    private void flower2_MouseMove(object sender, MouseEventArgs e)
    {
        if (e.Button == System.Windows.Forms.MouseButtons.Left)
        {
            flower2.Left = e.X + flower2.Left - MouseDownLocation.X;
            flower2.Top = e.Y + flower2.Top - MouseDownLocation.Y;
        }
    }

我想要它做的是单击图像，创建一个新图像并能够拖放它。只有当我将代码放在页面顶部时，我才成功。这不是我想要的，因为我希望能够添加尽可能多的图像。我尝试了很多不同的方法。错误发生在：

      flower2.Left = e.X + flower2.Left - MouseDownLocation.X;
      flower2.Top = e.Y + flower2.Top - MouseDownLocation.Y;

在flower2名下的所有单词。这是因为，我已经在pictureBox2_Click中定义了flower2，所以每次点击一个新的PictureBox都会生成。但是，我需要制作它，以便我可以生成任意数量的图像而不将其放在页面的顶部，这使得它一次只使用一个图像。

Answer 1

您制作的每个图片框都会路由到这些事件，因此sender对象就是图片框，您只需要将其转换为正确的类型然后移动它。

 PictureBox pb = (PictureBox)sender;
 pb.Left = e.X + pb.Left - MouseDownLocation.X;

C＃ - 拖放＆amp;保持控制：扩展

1 个答案: