我有一个PictureBox1,它的sizemode设置为Stretch和PictureBox1。 PictureBox1包含一个图像,让我们选择它的一部分,然后对其进行裁切并将裁切后的部分存储在PictureBox2中。 当sizemode设置为“拉伸”且图片未缩放时,效果很好。,但当我缩放它或将sizemode设置为zoom时,效果不佳。
working example - sizemode set to 'stretch'
我用来裁剪部分图片的代码(original source)
for character in word:
if chr(Alphabet[Alphabet.find(character)) >= "z":
new_Alphabet = Alphabet.join(Alphabet)
elif chr(Alphabet[Alphabet.find(character)) <= "a":
new_Alphabet = Alphabet.join(Alphabet[:-1])
如何正确计算?如何使裁剪功能发挥作用,使用户可以放大/缩小并仍然裁剪图片的正确部分?
答案 0 :(得分:-1)
您需要使用拉伸系数以及偏移量来计算点。
对于Zoom
,只有一个因素,因为Image
和PictureBox
的纵横比始终相同,但是通常存在偏移;对于Stretch
,您不需要补偿,但两个因子。
下面是一个使用两个 PictureBoxes
的示例,其中两个显示了缩放版本和裁剪后的位图。它利用确定大小和偏移量的通用功能ImageArea
。
两个类级变量:
Point pDown = Point.Empty;
Rectangle rect = Rectangle.Empty;
三个鼠标事件:
private void PictureBox1_MouseDown(object sender, MouseEventArgs e)
{
pDown = e.Location;
pictureBox1.Refresh();
}
private void PictureBox1_MouseMove(object sender, MouseEventArgs e)
{
if (!e.Button.HasFlag(MouseButtons.Left)) return;
rect = new Rectangle(pDown, new Size(e.X - pDown.X, e.Y - pDown.Y));
using (Graphics g = pictureBox2.CreateGraphics())
{
pictureBox1.Refresh();
g.DrawRectangle(Pens.Orange, rect);
}
}
private void PictureBox1_MouseUp(object sender, MouseEventArgs e)
{
Rectangle iR = ImageArea(pictureBox2);
rect = new Rectangle(pDown.X - iR.X, pDown.Y - iR.Y,
e.X - pDown.X, e.Y - pDown.Y);
Rectangle rectSrc = Scaled(rect, pictureBox2, true);
Rectangle rectDest = new Rectangle(Point.Empty, rectSrc.Size);
Bitmap bmp = new Bitmap(rectDest.Width, rectDest.Height);
using (Graphics g = Graphics.FromImage(bmp))
{
g.DrawImage(pictureBox1.Image, rectDest, rectSrc, GraphicsUnit.Pixel);
}
pictureBox2.Image = bmp;
}
这是一个有用的函数,可针对任何sizemode返回图片框内的实际图像区域。.
Rectangle ImageArea(PictureBox pbox)
{
Size si = pbox.Image.Size;
Size sp = pbox.ClientSize;
if (pbox.SizeMode == PictureBoxSizeMode.StretchImage)
return pbox.ClientRectangle;
if (pbox.SizeMode == PictureBoxSizeMode.Normal ||
pbox.SizeMode == PictureBoxSizeMode.AutoSize)
return new Rectangle(Point.Empty, si);
if (pbox.SizeMode == PictureBoxSizeMode.CenterImage)
return new Rectangle(new Point((sp.Width - si.Width) / 2,
(sp.Height - si.Height) / 2), si);
// PictureBoxSizeMode.Zoom
float ri = 1f * si.Width / si.Height;
float rp = 1f * sp.Width / sp.Height;
if (rp > ri)
{
int width = si.Width * sp.Height / si.Height;
int left = (sp.Width - width) / 2;
return new Rectangle(left, 0, width, sp.Height);
}
else
{
int height = si.Height * sp.Width / si.Width;
int top = (sp.Height - height) / 2;
return new Rectangle(0, top, sp.Width, height);
}
}
我们只需要偏移即可确定未缩放的矩形。我们还需要扩展它:
Rectangle Scaled(Rectangle rect, PictureBox pbox, bool scale)
{
float factor = GetFactor(pbox);
if (!scale) factor = 1f / factor;
return Rectangle.Round(new RectangleF(rect.X * factor, rect.Y * factor,
rect.Width * factor, rect.Height * factor));
}
为此,需要知道缩放比例,具体取决于长宽比:
float GetFactor(PictureBox pBox)
{
if (pBox.Image == null) return 0;
Size si = pBox.Image.Size;
Size sp = pBox.ClientSize;
float ri = 1f * si.Width / si.Height;
float rp = 1f * sp.Width / sp.Height;
float factor = 1f * pBox.Image.Width / pBox.ClientSize.Width;
if (rp > ri) factor = 1f * pBox.Image.Height / pBox.ClientSize.Height;
return factor;
}
如果通过将PictureBox
放在AutoScrolling Panel
内并更改Pbox.Size
来放大或缩小,此解决方案也将起作用。