如何歪斜圖像？例如，每個角都有一個CGPoint，坐標為p1，p2，p3，p4。然后，我需要設置-p4.x + = 50，p4.y + = 30。因此，應在2D透視圖中拉伸此角（p4），并使圖像變形。替代文字（來源：polar-b.com）我嘗試使用CATransform3D，但是似乎不能以這種方式完成，因為這只是改變視角（旋轉，使一側更近/更遠）。也許CGAffineTransform有用嗎？如果您知道答案，請編寫示例代碼。
3 回答
慕田峪7331174
TA貢獻1828條經驗 獲得超13個贊
我在Swift中嘗試了一個很棒的答案@KennyTM，并收到一個錯誤“表達式太復雜，無法在合理的時間內解決”。
因此，這是Swift的簡化版本：
let y21 = y2a - y1a
let y32 = y3a - y2a
let y43 = y4a - y3a
let y14 = y1a - y4a
let y31 = y3a - y1a
let y42 = y4a - y2a
let a = -H*(x2a*x3a*y14 + x2a*x4a*y31 - x1a*x4a*y32 + x1a*x3a*y42)
let b = W*(x2a*x3a*y14 + x3a*x4a*y21 + x1a*x4a*y32 + x1a*x2a*y43)
let c0 = -H*W*x1a*(x4a*y32 - x3a*y42 + x2a*y43)
let cx = H*X*(x2a*x3a*y14 + x2a*x4a*y31 - x1a*x4a*y32 + x1a*x3a*y42)
let cy = -W*Y*(x2a*x3a*y14 + x3a*x4a*y21 + x1a*x4a*y32 + x1a*x2a*y43)
let c = c0 + cx + cy
let d = H*(-x4a*y21*y3a + x2a*y1a*y43 - x1a*y2a*y43 - x3a*y1a*y4a + x3a*y2a*y4a)
let e = W*(x4a*y2a*y31 - x3a*y1a*y42 - x2a*y31*y4a + x1a*y3a*y42)
let f0 = -W*H*(x4a*y1a*y32 - x3a*y1a*y42 + x2a*y1a*y43)
let fx = H*X*(x4a*y21*y3a - x2a*y1a*y43 - x3a*y21*y4a + x1a*y2a*y43)
let fy = -W*Y*(x4a*y2a*y31 - x3a*y1a*y42 - x2a*y31*y4a + x1a*y3a*y42)
let f = f0 + fx + fy;
let g = H*(x3a*y21 - x4a*y21 + (-x1a + x2a)*y43)
let h = W*(-x2a*y31 + x4a*y31 + (x1a - x3a)*y42)
let i0 = H*W*(x3a*y42 - x4a*y32 - x2a*y43)
let ix = H*X*(x4a*y21 - x3a*y21 + x1a*y43 - x2a*y43)
let iy = W*Y*(x2a*y31 - x4a*y31 - x1a*y42 + x3a*y42)
var i = i0 + ix + iy
let epsilon = CGFloat(0.0001);
if fabs(i) < epsilon {
i = epsilon * (i > 0 ? 1 : -1);
}
return CATransform3D(m11: a/i, m12: d/i, m13: 0, m14: g/i, m21: b/i, m22: e/i, m23: 0, m24: h/i, m31: 0, m32: 0, m33: 1, m34: 0, m41: c/i, m42: f/i, m43: 0, m44: 1.0)
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