Question

我想同时显示四个图像，并在表单加载时切换图像。目前，图像将以不同的数字显示，例如：将出现1个图像或2个图像，等等4.我还要确保不会出现重复。

Form1_Load中的代码：

PictureBox[] boxes = new PictureBox[4];

boxes[0] = pictureBox0;
boxes[1] = pictureBox1;
boxes[2] = pictureBox2;
boxes[3] = pictureBox3;

for (int i = 0; i < boxes.Length; i++)
{
    int switcher = r.Next(0, 5);

    switch (switcher)
    {
        case 0:
            { boxes[i].Image = Properties.Resources.dog0; } break;
        case 1:
            { boxes[i].Image = Properties.Resources.dog1; } break;
        case 2:
            { boxes[i].Image = Properties.Resources.dog2; } break;
        case 3:
            { boxes[i].Image = Properties.Resources.dog3; } break;
    }
}

上面给出了关于当前发生的事情的两个例子。

更新 - 工作

程序现在在加载时移动图像并且没有重复：）

List<Bitmap> resources = new List<Bitmap>();
resources.Add(Properties.Resources.dog0);
resources.Add(Properties.Resources.dog1);
resources.Add(Properties.Resources.dog2);
resources.Add(Properties.Resources.dog3);

resources = resources.OrderBy(a => Guid.NewGuid()).ToList();

for (int i = 0; i < resources.Count; i++)
{
    pictureBox0.Image = resources[0];
    pictureBox1.Image = resources[1];
    pictureBox2.Image = resources[2];
    pictureBox3.Image = resources[3];
}

上面给出的两个例子显示了它现在发生了什么。

Answer 1

正如M.kazem Ahkhary指出你需要改变图像：

List<Bitmap> resources = new List<Bitmap>();
resources.Add(Properties.Resources.dog0);
resources.Add(Properties.Resources.dog1);
resources.Add(Properties.Resources.dog2);
resources.Add(Properties.Resources.dog3);

resources = resources.OrderBy(a => Guid.NewGuid()).ToList(); // Dirty but effective shuffle method

pictureBox0.Image = resources[0];
pictureBox1.Image = resources[1];
pictureBox2.Image = resources[2];
pictureBox3.Image = resources[3];

Answer 2

实施非常简单。首先，您需要对数组进行混洗，然后迭代它。 Fisher–Yates shuffle

创建方法ShuffleImages，如下所示：

public void ShuffleImages(PictureBox[] img)
{
    Random r = new Random();
    for (int i = 0; i < img.Length - 1; i++)
    {
        int j = r.Next(i, img.Length);
        PictureBox temp = img[j];
        img[j] = img[i];
        img[i] = temp;                
    }
}

并在Form1_Load事件中调用该方法：

private void Form1_Load(object sender, EventArgs e)
{
    PictureBox[] boxes = new PictureBox[4];
    boxes[0] = pictureBox0;
    boxes[1] = pictureBox1;
    boxes[2] = pictureBox2;
    boxes[3] = pictureBox3;

    ShuffleImages(boxes); //call the method

    for (int i = 0; i <= 3; i++)
    {
        switch (i)
        {
            case 0:
                {  boxes[i].Image = Properties.Resources.dog0;  }
                break;
            case 1:
                {  boxes[i].Image = Properties.Resources.dog1;  }
                break;
            case 2:
                {  boxes[i].Image = Properties.Resources.dog2;  }
                break;
            case 3:
                {  boxes[i].Image = Properties.Resources.dog3;  }
                break;
        }
    }
}

通过多个图片框循环多个图像

2 个答案: