Can jQuery get all CSS styles associated with an element?

Is there a way in jQuery to get all CSS from an existing element and apply it to another without listing them all?

I know it would work if they were a style attribute with attr(), but all of my styles are in an external style sheet.

Solution 1:

A couple years late, but here is a solution that retrieves both inline styling and external styling:

function css(a) {
    var sheets = document.styleSheets, o = {};
    for (var i in sheets) {
        var rules = sheets[i].rules || sheets[i].cssRules;
        for (var r in rules) {
            if (a.is(rules[r].selectorText)) {
                o = $.extend(o, css2json(rules[r].style), css2json(a.attr('style')));
            }
        }
    }
    return o;
}

function css2json(css) {
    var s = {};
    if (!css) return s;
    if (css instanceof CSSStyleDeclaration) {
        for (var i in css) {
            if ((css[i]).toLowerCase) {
                s[(css[i]).toLowerCase()] = (css[css[i]]);
            }
        }
    } else if (typeof css == "string") {
        css = css.split("; ");
        for (var i in css) {
            var l = css[i].split(": ");
            s[l[0].toLowerCase()] = (l[1]);
        }
    }
    return s;
}

Pass a jQuery object into css() and it will return an object, which you can then plug back into jQuery's $().css(), ex:

var style = css($("#elementToGetAllCSS"));
$("#elementToPutStyleInto").css(style);

:)

Solution 2:

Two years late, but I have the solution you're looking for. Not intending to take credit form the original author, here's a plugin which I found works exceptionally well for what you need, but gets all possible styles in all browsers, even IE.

Warning: This code generates a lot of output, and should be used sparingly. It not only copies all standard CSS properties, but also all vendor CSS properties for that browser.

jquery.getStyleObject.js:

/*
 * getStyleObject Plugin for jQuery JavaScript Library
 * From: http://upshots.org/?p=112
 */

(function($){
    $.fn.getStyleObject = function(){
        var dom = this.get(0);
        var style;
        var returns = {};
        if(window.getComputedStyle){
            var camelize = function(a,b){
                return b.toUpperCase();
            };
            style = window.getComputedStyle(dom, null);
            for(var i = 0, l = style.length; i < l; i++){
                var prop = style[i];
                var camel = prop.replace(/\-([a-z])/g, camelize);
                var val = style.getPropertyValue(prop);
                returns[camel] = val;
            };
            return returns;
        };
        if(style = dom.currentStyle){
            for(var prop in style){
                returns[prop] = style[prop];
            };
            return returns;
        };
        return this.css();
    }
})(jQuery);

Basic usage is pretty simple, but he's written a function for that as well:

$.fn.copyCSS = function(source){
  var styles = $(source).getStyleObject();
  this.css(styles);
}

Hope that helps.

Solution 3:

Why not use .style of the DOM element? It's an object which contains members such as width and backgroundColor.

Solution 4:

I had tried many different solutions. This was the only one that worked for me in that it was able to pick up on styles applied at class level and at style as directly attributed on the element. So a font set at css file level and one as a style attribute; it returned the correct font.

It is simple! (Sorry, can't find where I originally found it)

//-- html object
var element = htmlObject; //e.g document.getElementById
//-- or jquery object
var element = htmlObject[0]; //e.g $(selector)

var stylearray = document.defaultView.getComputedStyle(element, null);
var font = stylearray["font-family"]

Alternatively you can list all the style by cycling through the array

for (var key in stylearray) {
console.log(key + ': ' + stylearray[key];
}