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jQuery .position() strangeness while using CSS3 rotate attribute

I'm getting absolutely positioned rotated elements position with jQuery .position() method, then setting position-related attributes (top, left) with jQuery .css(pos), where pos is the data returned by .position(). I think it'll leave the element in it's place, but elements position is changing.

How can I use set rotated elements position, so that it'll be placed as expected? Maybe there is a coefficient depended on angle that changes position?

I'm testing in Google Chrome v.9, Windows XP.

HTML

<div id="container">
  <div id="element"> 
    <img src="http://t0.gstatic.com/images?q=tbn:ANd9GcS0Fawya9MVMez80ZusMVtk_4-ScKCIy6J_fg84oZ37GzKaJXU74Ma0vENc"/>
  </div>
</div> 

CSS

#container {
    position: relative;   
    border: 1px solid #999;
    padding: 5px;
    height: 300px;
    width:300px;
}    
#element {
    position: absolute;
    top:50px;
    left: 60px;
    width: auto;
    border: 1px solid #999;
    padding: 5px;
    -webkit-transform: rotate(45deg);
    -moz-transform: rotate(45deg);
}

JS

$(document).ready(function(){
    var $el = $('#element'),
    // getting position    
        pos = $el.position();
    alert(pos.left + '/' + pos.top);
    // alerts 37/11

    // setting css position attributes equal to pos
    $el.css(pos);
    // re-getting position
    pos = $el.position();
    alert(pos.left + '/' + pos.top);
    // alerts 14/-28
});    

View it http://jsfiddle.net/Antaranian/2gVL4/

like image 403
Mushex Antaranian Avatar asked Feb 22 '11 13:02

Mushex Antaranian


People also ask

What does position () method do in jQuery?

jQuery position() Method The position() method returns the position (relative to its parent element) of the first matched element. This method returns an object with 2 properties; the top and left positions in pixels.

How do you rotate something in jQuery?

rotate(angle) Using jQuery selector all included images are rotated. Example: $("#img"). rotate(45);

How do you rotate in JavaScript?

To rotate an image with JavaScript, access the image element with a method like getElementById() , then set the style. transform property to a string in the format rotate({value}deg) , where {value} is the clockwise angle of rotation in degrees. rotated.


2 Answers

// 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.

like image 179
Šime Vidas Avatar answered Oct 14 '22 14:10

Šime Vidas


Had similar problem. There is simple solution (not elegant, but working):

  • set current angle to 0
  • read X/Y position
  • revert angle back to its value

    var temp = $(this).position();
    temp.angle = getRotationDegrees( $(this) );            // remember current angle
    rotateObject($(this), 0);                              // set angle to 0
    temp.left = Math.round($(this).position().left);       // proper value
    temp.top = Math.round($(this).position().top);         // proper value
    
    // revert back the angle
    rotateObject($(this), temp.angle); 
    

Used functions:

    function rotateObject(obj, angle) {
        obj.css({ '-webkit-transform': 'rotate(' + angle + 'deg)'});
        obj.css({ '-moz-transform': 'rotate(' + angle + 'deg)'});
        obj.css({ '-ms-transform': 'rotate(' + angle + 'deg)'});
        obj.css({ 'msTransform': 'rotate(' + angle + 'deg)'});
        obj.css({ '-o-transform': 'rotate(' + angle + 'deg)'});
        obj.css({ '-sand-transform': 'rotate(' + angle + 'deg)'});
        obj.css({ 'transform': 'rotate(' + angle + 'deg)'});
    }

    function getRotationDegrees(obj) {
        var matrix = obj.css("-webkit-transform") ||
        obj.css("-moz-transform")    ||
        obj.css("-ms-transform")     ||
        obj.css("-o-transform")      ||
        obj.css("transform");
        if(matrix !== 'none') {
            var tr;
            if (tr = matrix.match('matrix\\((.*)\\)')) {
                tr = tr[1].split(',');
                if(typeof tr[0] != 'undefined' && typeof tr[1] != 'undefined') {
                    var angle = Math.round(Math.atan2(tr[1], tr[0]) * (180/Math.PI));
                }else{
                    var angle = 0; 
                }
            }else if(tr = matrix.match('rotate\\((.*)deg\\)')){
                var angle = parseInt(tr[1]); 
            }
        } else { var angle = 0; }
        return (angle < 0) ? angle + 360 : angle;
    }
like image 28
Jaroslaw Padula Avatar answered Oct 14 '22 14:10

Jaroslaw Padula