jQuery .position() strangeness while using CSS3 rotate attribute

Question

I\'m getting absolutely positioned rotated elements position with jQuery .position() method, then setting position-related attributes (top, left) with jQuery

Answer 1

// Needed to read the "real" position
$.fn.adjustedPosition = function() {
    var p = $(this).position();
    return {
        left: p.left - this.data('dx'),
        top: p.top - this.data('dy')
    }
};

---

$(function() { 

    var img = $('img'),
        pos;

    // Calculate the delta
    img.each(function() {
        var po = $(this).position(), // original position
            pr = $(this).addClass('rot').position(); // rotated position

        $(this).data({
            dx: pr.left - po.left, // delta X
            dy: pr.top - po.top // delta Y
        });
    });

    // Read the position
    pos = img.adjustedPosition();    
    alert(pos.left + '/' + pos.top);     

    // Write the position
    img.css(pos);

    // Read the position again
    pos = img.adjustedPosition();    
    alert(pos.left + '/' + pos.top);

});

Live demo: http://jsfiddle.net/2gVL4/4/

So what is going on here:

1. The CSS code that rotates the image is stored inside a special CSS class. I do this because I want to read the original position of the image (before rotating). Once I read that original position, I apply the .rot class, and then read the position again to calculate the difference (delta), which is stored inside the element's data().

2. Now, I can read the position via the custom method adjustedPosition (which is defined above). This method will read the position of the element and then subtract the delta values stored inside the data() of the element.

3. To write the position, just use the css(pos) method like normally.