HTML CSS3 / Canvas图像失真

Question

有没有办法在HTML5中使用CSS3或canvas标签来扭曲/扭曲一个角？

以下是Photoshop教程的截图如何操作：

更新

这是迄今为止我发现的最好的，但它不是100％准确： https://github.com/edankwan/PerspectiveTransform.js

UPDATE2：

我需要html5版本： http://www.rubenswieringa.com/code/as3/flex/DistortImage/

Answer 1

这应该对你有所帮助。 Link1

您应该在发布问题之前尝试搜索。我搜索了 html5画布倾斜图像，它向我展示了这么多结果。

<小时/> 的更新
查看此Fiddle

// Find each img, and replace it with a canvas
$('img').each(function (index, el) {
var c,      // new canvas which will replace this img element
    ctx,    // context of new canvas
    i,      // loop counter
    tmpCtx, // temp context for doing work
    h,      // height of the image / new canvas
    w,      // width of the image / new canvas
    dh,     // destination height (used in translation)
    dw,     // destination width (used in translation)
    dy,     // destination y
    leftTop,// left top corner position
    leftBot;// left bottom corner position

// Get the height/width of the image
h = el.height;
w = el.width;

// Create the canvas and context that will replace the image
c = $("<canvas height='" + h + "' width='" + w + "'><\/canvas>");
ctx = c.get(0).getContext('2d');

// Create a temporary work area
tmpCtx = document.createElement('canvas').getContext('2d');

// Draw the image on the temp work area
for (i = 0; i < h; i++) {
    dw = Math.abs((w * (h - i) + w * i) / h);
    tmpCtx.drawImage(el,
        0, i, w, 1,   // sx, sy, sw, sh
        0, i, dw, 1); // dx, dy, dw, dh
}

// Calculate the left corners to be 20% of the height
leftTop = parseInt(h * .2, 10);
leftBot = parseInt(h * 1, 10) - leftTop;

ctx.save();

// Draw the image on our real canvas
for (i = 0; i < w; i++) {
    dy = (leftTop * (w - i)) / w;
    dh = (leftBot * (w - i) + h * i) / w;
    ctx.drawImage(tmpCtx.canvas,
        i, 0, 1, h,
        i, dy, 1, dh);
}

ctx.restore();

// Replace the image with the canvas version
$(el).replaceWith(c);
});

Answer 2

我认为this正是您所寻找的。