图形转动
如何在vb6.0中实现图形转动 问题点数：0、回复次数：6Top
1 楼BitBlt（Raster Operater）回复于 2004-05-04 14:10:26 得分 0
任意角度旋转图形
来源:cww VB5.0进阶程式开发指南
如果有一个图范围如下：
P1+----------+ P2
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P3+----------+ P4
我们以P3为轴，顺时针转某个角度，如果是NT的使用者，可以使用PlgBlt()这个API来完
成，如果是Win95那就没办法了，这里也提供 VB5.0进阶程式开发指南所用的方法来完成
PlgBlt()的第二个叁数是重点，它是指向POINTAPI的一个3个元素之阵列，代表图形旋转
之後的三个端点，也就是说顺时针转了某个角度後，P1->P1' P2->P2' P3->P3'，这阵
列便是存P1' P2' P3'。 所以了，重点在於这三个点的取得，数学不好的我可真痛苦；
如果这三个点没有设好，那会产生一个变形的图，原本 PlgBlt()便是用来扭曲图形用的
，只是那三个点如果设好的话，刚好可以来旋转图。
PlgBlt(
HDC hdcDest, // handle of destination device context
CONST POINT * lpPoint, // vertices of destination parallelogram
HDC hdcSrc, // handle of source device context
int nXSrc, // x-coord. of upper-left corner of source rect.
int nYSrc, // y-coord. of upper-left corner of source rect.
int nWidth, // width of source rectangle
int nHeight, // height of source rectangle
HBITMAP hbmMask, // handle of bitmask
int xMask, // x-coord. of upper-left corner of bitmask rect.
int yMask // y-coord. of upper-left corner of bitmask
)
需2个Command Button 2个PictureBox，目的在Picture2画出旋转图
Option Explicit
Private Declare Function PlgBlt Lib "gdi32" (ByVal hdcDest As Long, _
lpPoint As POINTAPI, ByVal hdcSrc As Long, ByVal nXSrc As Long, _
ByVal nYSrc As Long, ByVal nWidth As Long, ByVal nHeight As Long, _
ByVal hbmMask As Long, ByVal xMask As Long, ByVal yMask As Long) As Long
Private Type POINTAPI
x As Long
y As Long
End Type
Const pi = 3.14159265358979
Dim dThetaDeg As Double
'以下只适用於NT
Private Sub Command1_Click()
Dim pt(1 To 3) As POINTAPI, p4 As POINTAPI
Dim dx As Long, dy As Long
Dim i As Long, offsetX As Long, offsetY As Long
Dim sida As Double
Dim MaxX As Long, MaxY As Long, MinX As Long, MinY As Long
sida = dThetaDeg * pi / 180
dx = Me.ScaleX(Picture1.Picture.Width, vbHimetric, vbPixels)
dy = Me.ScaleX(Picture1.Picture.Height, vbHimetric, vbPixels)
pt(1).x = dy * Sin(sida)
pt(1).y = dy - dy * Cos(sida)
pt(2).x = pt(1).x + dx * Cos(sida)
pt(2).y = pt(1).y + dx * Sin(sida)
pt(3).x = 0: pt(3).y = dy
'p4不用在pt()之阵列，它是由pt(1)-pt(3)所推出
p4.x = pt(3).x + dx * Cos(sida)
p4.y = pt(3).y + dx * Sin(sida)
'以下的工作只是为了让旋转後的图在Picture2的中间显示
MaxX = IIf(pt(1).x > pt(2).x, pt(1).x, pt(2).x)
MaxX = IIf(MaxX > pt(3).x, MaxX, pt(3).x)
MaxX = IIf(MaxX > p4.x, MaxX, p4.x)
MinX = IIf(pt(1).x < pt(2).x, pt(1).x, pt(2).x)
MinX = IIf(MinX < pt(3).x, MinX, pt(3).x)
MinX = IIf(MinX < p4.x, MinX, p4.x)
MaxY = IIf(pt(1).y > pt(2).y, pt(1).y, pt(2).y)
MaxY = IIf(MaxY > pt(3).y, MaxY, pt(3).y)
MaxY = IIf(MaxY > p4.y, MaxY, p4.y)
MinY = IIf(pt(1).y < pt(2).y, pt(1).y, pt(2).y)
MinY = IIf(MinY < pt(3).y, MinY, pt(3).y)
MinY = IIf(MinY < p4.y, MinY, p4.y)
offsetX = (Picture2.ScaleWidth - MaxX + MinX) \ 2 - MinX
offsetY = (Picture2.ScaleHeight - (MaxY - MinY)) \ 2 - MinY
pt(1).x = pt(1).x + offsetX: pt(1).y = pt(1).y + offsetY
pt(2).x = pt(2).x + offsetX: pt(2).y = pt(2).y + offsetY
pt(3).x = pt(3).x + offsetX: pt(3).y = pt(3).y + offsetY
i = PlgBlt(Picture2.hDC, pt(1), Picture1.hDC, 0, 0, dx, dy, 0, 0, 0)
End Sub
'这个方式不管95/NT都可行，来自vb5程式设计进阶指南一书，但速度慢
Private Sub Command2_Click()
Dim nX As Integer, nY As Integer
Dim nX1 As Integer, nY1 As Integer
Dim dX2 As Double, dY2 As Double
Dim dX3 As Double, dY3 As Double
Dim dThetaRad As Double
'Compute angle in radians
dThetaRad = -dThetaDeg * pi / 180
'Set scale modes to pixels
For nX = 0 To Picture2.ScaleWidth
nX1 = nX - Picture2.ScaleWidth \ 2
For nY = 0 To Picture2.ScaleHeight
nY1 = nY - Picture2.ScaleHeight \ 2
'Rotate picture by dThetaRad
dX2 = nX1 * Cos(-dThetaRad) + nY1 * Sin(-dThetaRad)
dY2 = nY1 * Cos(-dThetaRad) - nX1 * Sin(-dThetaRad)
'Translate to center of picture box
dX3 = dX2 + Picture1.ScaleWidth \ 2
dY3 = dY2 + Picture1.ScaleHeight \ 2
'If data point is in picOne, set its color in picTwo
If dX3 > 0 And dX3 < Picture1.ScaleWidth - 1 _
And dY3 > 0 And dY3 < Picture1.ScaleHeight - 1 Then
Picture2.PSet (nX, nY), Picture1.Point(dX3, dY3)
End If
Next nY
Next nX
End Sub
Private Sub Form_Load()
Picture1.ScaleMode = 3
Picture2.ScaleMode = 3
'设定旋转角度
dThetaDeg = 30
Set Picture1.Picture = LoadPicture("e:\cli.bmp")
End Sub
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2 楼ljf88888（feng）回复于 2004-05-05 12:03:28 得分 0
upTop
3 楼kmzs（.:RNPA:.山水岿濛）回复于 2004-05-05 12:09:01 得分 0
好用Top
4 楼hhyttppd（123456）回复于 2004-05-05 12:15:21 得分 0
除了前面的方法，当然你可以用图形学里旋转矩阵对它进行变换。不过就是太慢，闪烁太大，也可能实现不了动画效果。
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5 楼thirdapple（.:RNPA:.陨落雕 - 芝兰宝树）回复于 2004-05-05 12:37:57 得分 0
呵呵，To 楼上的，都是用旋转矩阵实现的……Top
6 楼yangchonglv（瓯越浪子）回复于 2004-05-05 13:52:07 得分 0
有没有只转动90、270读的，那比较简单Top
