14

我有动画 gif，我正在使用一个类来解析其中的图像（帧）。课程是：

using System;
using System.Drawing;
using System.Drawing.Imaging;
using System.Collections.Generic;
using System.IO;

public class AnimatedGif
{
    private List<AnimatedGifFrame> mImages = new List<AnimatedGifFrame>();
    public AnimatedGif(string path)
    {
        Image img = Image.FromFile(path);
        int frames = img.GetFrameCount(FrameDimension.Time);
        if (frames <= 1) throw new ArgumentException("Image not animated");
        byte[] times = img.GetPropertyItem(0x5100).Value;
        int frame = 0;
        for (; ; )
        {
            int dur = BitConverter.ToInt32(times, 4 * frame);
            mImages.Add(new AnimatedGifFrame(new Bitmap(img), dur));
            if (++frame >= frames) break;
            img.SelectActiveFrame(FrameDimension.Time, frame);
        }
        img.Dispose();
    }
    public List<AnimatedGifFrame> Images { get { return mImages; } }
}

public class AnimatedGifFrame
{
    private int mDuration;
    private Image mImage;
    internal AnimatedGifFrame(Image img, int duration)
    {
        mImage = img; mDuration = duration;
    }
    public Image Image { get { return mImage; } }
    public int Duration { get { return mDuration; } }
}

现在在form1中，我在这种情况下循环帧4，我想将动画旋转任意角度。现在它每旋转 45 或 90 度。我想为动画添加更多帧（图像），所以如果我将旋转设置为 31 或 10 度，我会看到动画旋转 10 度。

这是 Form1 中的代码，效果不好。我正在使用一个我尚未测试的旋转功能，如果它有任何工作。

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;

namespace AnimatedGifEditor
{
    public partial class Form1 : Form
    {
        Image myImage;
        AnimatedGif myGif;
        Bitmap bitmap;

        public Form1()
        {
            InitializeComponent();

            myImage = Image.FromFile(@"D:\fananimation.gif");
            myGif = new AnimatedGif(@"D:\fananimation.gif");
            for (int i = 0; i < myGif.Images.Count; i++)
            {
                pictureBox1.Image = myGif.Images[3].Image;
                bitmap = new Bitmap(pictureBox1.Image);
                rotateImage(bitmap, 76);
                pictureBox1.Image = bitmap;
            }
        }

        private void Form1_Load(object sender, EventArgs e)
        {

        }

private Bitmap RotateImg(Bitmap bmp, float angle, Color bkColor)
    {
        int w = bmp.Width;
        int h = bmp.Height;
        bmp.PixelFormat pf = default(bmp.PixelFormat);
        if (bkColor == Color.Transparent)
        {
            pf = bmp.Format32bppArgb;
        }
        else
        {
            pf = bmp.PixelFormat;
        }

        Bitmap tempImg = new Bitmap(w, h, pf);
        Graphics g = Graphics.FromImage(tempImg);
        g.Clear(bkColor);
        g.DrawImageUnscaled(bmp, 1, 1);
        g.Dispose();

        GraphicsPath path = new GraphicsPath();
        path.AddRectangle(new RectangleF(0f, 0f, w, h));
        Matrix mtrx = new Matrix();
        //Using System.Drawing.Drawing2D.Matrix class 
        mtrx.Rotate(angle);
        RectangleF rct = path.GetBounds(mtrx);
        Bitmap newImg = new Bitmap(Convert.ToInt32(rct.Width), Convert.ToInt32(rct.Height), pf);
        g = Graphics.FromImage(newImg);
        g.Clear(bkColor);
        g.TranslateTransform(-rct.X, -rct.Y);
        g.RotateTransform(angle);
        g.InterpolationMode = InterpolationMode.HighQualityBilinear;
        g.DrawImageUnscaled(tempImg, 0, 0);
        g.Dispose();
        tempImg.Dispose();
        return newImg;
    }
    }
}

我用于测试的动画 gif 可以在这里找到：

在此处输入图像描述

4

7 回答 7

28

我不明白你的问题是什么，但我认为你的代码可以改进。我认为您不需要直接使用Matrix该类。有一些功能可以为您工作。事实上，您唯一需要的是：将旋转点设置为中心，旋转图形并在其上绘制，使用Graphics类的一些函数。因此，要旋转图像，您可以使用以下简单代码：

private Bitmap RotateImage(Bitmap bmp, float angle) {
     Bitmap rotatedImage = new Bitmap(bmp.Width, bmp.Height);
     rotatedImage.SetResolution(bmp.HorizontalResolution, bmp.VerticalResolution);

     using (Graphics g = Graphics.FromImage(rotatedImage)) {
        // Set the rotation point to the center in the matrix
        g.TranslateTransform(bmp.Width / 2, bmp.Height / 2);
        // Rotate
        g.RotateTransform(angle);
        // Restore rotation point in the matrix
        g.TranslateTransform(- bmp.Width / 2, - bmp.Height / 2);
        // Draw the image on the bitmap
        g.DrawImage(bmp, new Point(0, 0));
     }

     return rotatedImage;
}
于 2012-08-19T10:47:40.283 回答
9

根据之前的答案，我创建了不会剪切图像的代码（其他示例对我不起作用）

    private Bitmap RotateImage(Bitmap bmp, float angle)
    {
        float height = bmp.Height;
        float width = bmp.Width;
        int hypotenuse = System.Convert.ToInt32(System.Math.Floor(Math.Sqrt(height * height + width * width)));
        Bitmap rotatedImage = new Bitmap(hypotenuse, hypotenuse);
        using (Graphics g = Graphics.FromImage(rotatedImage))
        {
            g.TranslateTransform((float)rotatedImage.Width / 2, (float)rotatedImage.Height / 2); //set the rotation point as the center into the matrix
            g.RotateTransform(angle); //rotate
            g.TranslateTransform(-(float)rotatedImage.Width / 2, -(float)rotatedImage.Height / 2); //restore rotation point into the matrix
            g.DrawImage(bmp, (hypotenuse - width) / 2, (hypotenuse - height) / 2, width, height);
        }
        return rotatedImage;
    }
于 2017-08-17T09:58:52.867 回答
7

我自己尝试了@Omar 的答案并意识到，原始图像在两侧被切割......我已经重写了它，因此它将图像调整为新尺寸：

private static Bitmap RotateImage(Bitmap bmp, float angle)
{
    float alpha = angle;

    //edit: negative angle +360
    while(alpha <0) alpha +=360;

    float gamma = 90;
    float beta = 180 - angle - gamma;

    float c1 = bmp.Height;
    float a1 = (float)(c1 * Math.Sin(alpha * Math.PI / 180) / Math.Sin(gamma * Math.PI / 180));
    float b1 = (float)(c1 * Math.Sin(beta * Math.PI / 180) / Math.Sin(gamma * Math.PI / 180));

    float c2 = bmp.Width;
    float a2 = (float)(c2 * Math.Sin(alpha * Math.PI / 180) / Math.Sin(gamma * Math.PI / 180));
    float b2 = (float)(c2 * Math.Sin(beta * Math.PI / 180) / Math.Sin(gamma * Math.PI / 180));

    int width = Convert.ToInt32(b2 + a1);
    int height = Convert.ToInt32(b1 + a2);

    Bitmap rotatedImage = new Bitmap(width, height);
    using (Graphics g = Graphics.FromImage(rotatedImage))
    {
        g.TranslateTransform(rotatedImage.Width / 2, rotatedImage.Height / 2); //set the rotation point as the center into the matrix
        g.RotateTransform(angle); //rotate
        g.TranslateTransform(-rotatedImage.Width / 2, -rotatedImage.Height / 2); //restore rotation point into the matrix
        g.DrawImage(bmp, new Point((width - bmp.Width) / 2, (height - bmp.Height) / 2)); //draw the image on the new bitmap
    }
    return rotatedImage;
}

30 度半径处的尺寸变化

于 2017-07-05T17:06:21.267 回答
6

你试过旋转翻转吗？

public partial class Form1 : Form
{
    Image myImage;
    AnimatedGif myGif;
    Bitmap bitmap;
    public Form1()
    {
        InitializeComponent();
        myImage = Image.FromFile(@"D:\fananimation.gif");
        bitmap = new Bitmap(myImage);
        bitmap.RotateFlip(System.Drawing.RotateFlipType.Rotate90FlipNone);
        this.pictureBox1.Image = bitmap;
    }

}

来源

于 2012-08-19T06:37:15.447 回答
1

我在VB中使用这个函数：

    Public Function RotateImage(ByRef image As Image, ByVal angle As Single) As Drawing.Bitmap
    If image Is Nothing Then
        Throw New ArgumentNullException("image")
    End If

    Dim pi2 As Single = Math.PI / 2.0
    Dim oldWidth As Single = image.Width
    Dim oldHeight As Single = image.Height

    Dim theta As Single = angle * Math.PI / 180.0
    Dim locked_theta As Single = theta

    If locked_theta < 0.0 Then locked_theta += 2 * Math.PI

    Dim newWidth, newHeight As Single
    Dim nWidth, nHeight As Integer

    Dim adjacentTop, oppositeTop As Single
    Dim adjacentBottom, oppositeBottom As Single

    If (locked_theta >= 0.0 And locked_theta < pi2) Or _
    (locked_theta >= Math.PI And locked_theta < (Math.PI + pi2)) Then
        adjacentTop = Math.Abs(Math.Cos(locked_theta)) * oldWidth
        oppositeTop = Math.Abs(Math.Sin(locked_theta)) * oldWidth

        adjacentBottom = Math.Abs(Math.Cos(locked_theta)) * oldHeight
        oppositeBottom = Math.Abs(Math.Sin(locked_theta)) * oldHeight
    Else
        adjacentTop = Math.Abs(Math.Sin(locked_theta)) * oldHeight
        oppositeTop = Math.Abs(Math.Cos(locked_theta)) * oldHeight

        adjacentBottom = Math.Abs(Math.Sin(locked_theta)) * oldWidth
        oppositeBottom = Math.Abs(Math.Cos(locked_theta)) * oldWidth
    End If



    newWidth = adjacentTop + oppositeBottom
    newHeight = adjacentBottom + oppositeTop

    nWidth = Int(Math.Ceiling(newWidth))
    nHeight = Int(Math.Ceiling(newHeight))

    Dim rotatedBmp As New Drawing.Bitmap(nWidth, nHeight)

    Dim g As Graphics = Graphics.FromImage(rotatedBmp)

    Dim points(2) As Point

    If (locked_theta >= 0.0 And locked_theta < pi2) Then

        points(0) = New Point(Int(oppositeBottom), 0)
        points(1) = New Point(nWidth, Int(oppositeTop))
        points(2) = New Point(0, Int(adjacentBottom))

    ElseIf locked_theta >= pi2 And locked_theta < Math.PI Then

        points(0) = New Point(nWidth, Int(oppositeTop))
        points(1) = New Point(Int(adjacentTop), nHeight)
        points(2) = New Point(Int(oppositeBottom), 0)

    ElseIf locked_theta >= Math.PI And locked_theta < (Math.PI + pi2) Then

        points(0) = New Point(Int(adjacentTop), nHeight)
        points(1) = New Point(0, Int(adjacentBottom))
        points(2) = New Point(nWidth, Int(oppositeTop))

    Else

        points(0) = New Point(0, Int(adjacentBottom))
        points(1) = New Point(Int(oppositeBottom), 0)
        points(2) = New Point(Int(adjacentTop), nHeight)
    End If

    g.DrawImage(image, points)

    g.Dispose()
    image.Dispose()

    Return rotatedBmp

End Function
于 2012-08-19T11:15:57.850 回答
0

根据 Timo 的代码，我做了一些改进，通过改进，可以成功地给出负角（最高 -360）作为参数

private static Bitmap RotateImage(Bitmap bmp, float angle) 
    {
        float alpha = angle;

        //edit: negative angle +360
        while (alpha < 0) alpha += 360;

        float gamma = 90;
        float beta = 180 - angle - gamma;

        float c1 = bmp.Height;
        float a1 = Math.Abs((float)(c1 * Math.Sin(alpha * Math.PI / 180)));
        float b1 = Math.Abs((float)(c1 * Math.Sin(beta * Math.PI / 180)));

        float c2 = bmp.Width;
        float a2 = Math.Abs((float)(c2 * Math.Sin(alpha * Math.PI / 180)));
        float b2 = Math.Abs((float)(c2 * Math.Sin(beta * Math.PI / 180)));

        int width = Convert.ToInt32(b2 + a1);
        int height = Convert.ToInt32(b1 + a2);

        Bitmap rotatedImage = new Bitmap(width, height);
        using (Graphics g = Graphics.FromImage(rotatedImage))
        {
            g.TranslateTransform(rotatedImage.Width / 2, rotatedImage.Height / 2); //set the rotation point as the center into the matrix
            g.RotateTransform(angle); //rotate
            g.TranslateTransform(-rotatedImage.Width / 2, -rotatedImage.Height / 2); //restore rotation point into the matrix
            g.DrawImage(bmp, new Point((width - bmp.Width) / 2, (height - bmp.Height) / 2)); //draw the image on the new bitmap
        }
        return rotatedImage;
    }
于 2020-10-29T16:59:57.090 回答
0

我检查了答案，他们都至少存在以下问题之一：

  • 裁剪/不正确的居中
  • 不必要的保证金
  • 某些角度范围的错误
  • 不必要的复杂计算/代码

该解决方案可以处理任何角度（正、负、超过 360° 等）。没有裁剪或过多的边距。也没有内存泄漏。

public Bitmap RotateBitmap(Bitmap bmp, float angle)
{
    double radianAngle = angle / 180.0 * Math.PI;
    double cosA = Math.Abs(Math.Cos(radianAngle));
    double sinA = Math.Abs(Math.Sin(radianAngle));

    int newWidth = (int)(cosA * bmp.Width + sinA * bmp.Height);
    int newHeight = (int)(cosA * bmp.Height + sinA * bmp.Width);

    var rotatedBitmap = new Bitmap(newWidth, newHeight);
    rotatedBitmap.SetResolution(bmp.HorizontalResolution, bmp.VerticalResolution);
    
    using (Graphics g = Graphics.FromImage(rotatedBitmap))
    {
        g.TranslateTransform(rotatedBitmap.Width / 2, rotatedBitmap.Height / 2);
        g.RotateTransform(angle);
        g.TranslateTransform(-bmp.Width / 2, -bmp.Height / 2);
        g.DrawImage(bmp, new Point(0, 0));
    }

    bmp.Dispose();//Remove if you want to keep oryginal bitmap

    return rotatedBitmap;
}
于 2022-01-09T00:41:52.270 回答