Question

I want to get an element’s position relative to the window (fixed position).

Here’s what I’ve got so far:

(function ($) {
    $.fn.fixedPosition = function () {
        var offset = this.offset();
        var $doc = $(document);
        return {
            'x': offset.left - $doc.scrollLeft(),
            'y': offset.top - $doc.scrollTop()
        };
    };
})(jQuery);

$('#thumbnails img').click(function () {
    var pos = $(this).fixedPosition();
    console.log(pos);
});

But when I click a thumbnail, it appears to be off by about 10 pixels or so. i.e., it will give me negative values for y even when the top edge of the photo is about 5 pixels away from the top of my browser window.

Answer 1

Update:

Solution now depends on JSizes and a couple helper methods:

function Point(x, y) {
    return {
        'x': x,
        'y': y,
        'left': x,
        'top': y
    };
}

$.fn.outerOffset = function () {
    /// <summary>Returns an element's offset relative to its outer size; i.e., the sum of its left and top margin, padding, and border.</summary>
    /// <returns type="Object">Outer offset</returns>
    var margin = this.margin();
    var padding = this.padding();
    var border = this.border();
    return Point(
        margin.left + padding.left + border.left,
        margin.top + padding.top + border.top
    );
};


$.fn.fixedPosition = function () {
    /// <summary>Returns the "fixed" position of the element; i.e., the position relative to the browser window.</summary>
    /// <returns type="Object">Object with 'x' and 'y' properties.</returns>
    var offset = this.offset();
    var $doc = $(document);
    var bodyOffset = $(document.body).outerOffset();
    return Point(offset.left - $doc.scrollLeft() + bodyOffset.left, offset.top - $doc.scrollTop() + bodyOffset.top);
};