I need to get list of elements all css attributes. How can I do that ?
A:
For inline styles:
var styles = $("#someelement").attr("style");
From there, you should be able to split this string if you need to loop the styles.
To check individual styles, check the docs:
ScottE
2009-09-24 11:39:10
I need also attributes that are set in css
newbie
2009-09-24 11:43:39
if that is possible
newbie
2009-09-24 11:44:11
Perhaps explain what you're trying to accomplish in your question. You may be better off checking for individual styles, or using classes and using .hasClass() - gnarf has a link below otherwise
ScottE
2009-09-24 14:00:00
A:
$("#div1).css("background-color");
will return the background color property of an element with id div1.
Sorry, I don't know a way to get all the CSS attributes using jQuery.
rahul
2009-09-24 11:39:38
+4
A:
Copying the source from SO1004475 - jQuery CSS plugin that returns computed style of element to pseudo clone that element? - Please follow link and upvote there if you find it useful.
It seems ridiculous, but this is probably your best option - makes .css()
with no arguments get an object with all this stuff set.
jQuery.fn.css2 = jQuery.fn.css;
jQuery.fn.css = function() {
if (arguments.length) return jQuery.fn.css2.apply(this, arguments);
var attr = ['font-family','font-size','font-weight','font-style','color',
'text-transform','text-decoration','letter-spacing','word-spacing',
'line-height','text-align','vertical-align','direction','background-color',
'background-image','background-repeat','background-position',
'background-attachment','opacity','width','height','top','right','bottom',
'left','margin-top','margin-right','margin-bottom','margin-left',
'padding-top','padding-right','padding-bottom','padding-left',
'border-top-width','border-right-width','border-bottom-width',
'border-left-width','border-top-color','border-right-color',
'border-bottom-color','border-left-color','border-top-style',
'border-right-style','border-bottom-style','border-left-style','position',
'display','visibility','z-index','overflow-x','overflow-y','white-space',
'clip','float','clear','cursor','list-style-image','list-style-position',
'list-style-type','marker-offset'];
var len = attr.length, obj = {};
for (var i = 0; i < len; i++)
obj[attr[i]] = jQuery.fn.css2.call(this, attr[i]);
return obj;
}
gnarf
2009-09-24 12:04:56