HTML CSS3 / Canvas Image Distortion - html

Question

Is there a way to skew/distort only one corner using CSS3 or canvas tag in HTML5?
Here is a screenshot from Photoshop tutorial how to do it:
Update:
This is the best I have found so far, but it is not 100% accurate:
https://github.com/edankwan/PerspectiveTransform.js
Update2:
I need html5 version of this:
http://www.rubenswieringa.com/code/as3/flex/DistortImage/

Answer 1

This should help you .
Link1
And you should try searching before posting a question. I searched for html5 canvas skew image and it showed me so many results .
Update
Check Out this 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

I think this is what you are looking for.