通过多个图片框循环多个图像

Loop multiple images through multiple pictureboxes

我想同时显示四张图片,并在表单加载时切换图片位置。目前,图片会以不同的数量出现,例如:会出现1张图片或2张图片,等等最多4张。我也想确保不会出现重复的。

来自 Form1_Load 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;
    }
}

上面给出了两个关于当前发生的事情的例子。

更新 - 工作

程序现在会在加载时四处移动图像,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展示了现在它工作时会发生什么。

正如 M.kazem Ahkhary 指出的那样,您需要随机播放图像:

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];

实现非常简单。首先,您需要打乱数组,然后遍历它。 Fisher–Yates shuffle.

创建方法 ShuffleImages 如下:

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;                
    }
}

并在您的 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;
        }
    }
}