Question

我想为Panel中的PictureBox生成一个随机位置。我用C＃加载我的页面。这是我的PictureBox代码。我怎么能有一个随机的位置呢？

private void Form1_Load(object sender, EventArgs e)
{            
    PictureBox pictureboxtroll = new PictureBox();
    pictureboxtroll.Size = new System.Drawing.Size(70, 70);
    pictureboxtroll.BackColor = Color.Green;
    this.Controls.Add(pictureboxtroll);
}

Answer 1

您可以通过Location属性设置元素的位置：

// generate random position
Random random = new Random();
var viewRange = this.Size - pictureboxtroll.Size;
var left = random.Next(0, viewRange.Width);
var top = random.Next(0, viewRange.Height);

// set the random position
pictureboxtroll.Location = new Point(left, top);

Answer 2

如果你想移动pictureBox（X和Y坐标）的位置，你必须通过Location属性访问它，如下所示：

Random rand = new Random();
picturebox.Location.X = new Point(rand.Next(0,this.Widht));
pictureBox.Location.Y = new Point(rand.Next(0,this.Height));

如果您希望在首次加载Form1对象时执行该操作，则需要将该小代码段放在Form_Load事件中

面板上图片框的随机位置

2 个答案: