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.
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"));
$("#elementToPutStyleInto").css(style);
:)
Two years late, but I have the solution you're looking for. Not intending to take credit form the original author, here's a plugin which I found works exceptionally well for what you need, but gets all possible styles in all browsers, even IE.
Warning: This code generates a lot of output, and should be used sparingly. It not only copies all standard CSS properties, but also all vendor CSS properties for that browser.
jquery.getStyleObject.js:
/*
* getStyleObject Plugin for jQuery JavaScript Library
* From: http://upshots.org/?p=112
*/
(function($){
$.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.
Why not use .style of the DOM element? It's an object which contains members such as width and backgroundColor .