pass 2 objects to jquery.css()

Question

on my way to the shortest code on urff.

i have an object, where i make a feature detection. i then pass the result to a jquery.css() command. but i do have some other css rules, that i need to apply, too. i cant put them into the object, cause these rules differ from time to time. so i now pass two jquery.css() commands, but im trying to make one out of it, just for the sexyness..

var myObject  = (Modernizr.csstransforms ? {transform: 'translateX(123px)'} : {left: 123});
$("#elem").css(myObject).css({ sexyness: 0 });

while i try to make it something like this:

$("#elem").css(myObject, { sexyness: "100%" });

am i missing something, or is this as good as it gets?

You should avoid using css as much as possible. If you need it for dynamic values, so be it, but for static styles you should be using addClass, removeClass, and toggleClass. — zzzzBov
– zzzzBov, Commented May 7, 2013 at 17:17

Arun P Johny · Accepted Answer · 2013-05-07 17:11:43Z

4

You can use $.extend() to merge those two css property objects

$("#elem").css($.extend({}, myObject, { sexyness: "100%" }));

answered May 7, 2013 at 17:11

Arun P Johny

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Community · Accepted Answer · 2020-06-20 09:12:55Z

2

You need "glue" objects and use new object for set style. Looks like this:

fixed

$.extend(myObject, { sexyness: "100%" });
$("#elem").css(myObject);

edited Jun 20, 2020 at 9:12

CommunityBot

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answered May 7, 2013 at 17:11

Eugene

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user400654 · Accepted Answer · 2013-05-07 17:15:09Z

2

$("#elem").css($.extend({}, myObject, { sexyness: "100%" }));

jQuery.extend

edited May 7, 2013 at 17:15

user400654

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answered May 7, 2013 at 17:12

Barmar

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Comments

TonyH · Accepted Answer · 2013-05-07 18:30:12Z

0

Wrong :( Misread the API ~~jQuery 1.9 allows you to use arrays of properties, so you can do:~~

~~var myObject = (Modernizr.csstransforms ? ["transform: translateX(123px)"] : ["left: 123"]); $("#elem").css(myObject.add("sexyness: 80"))~~

Lets cheat and use JSON.parse, which lets us simply concatenate strings

var myObject  = (Modernizr.csstransforms ? 
                 'transform: "translateX(123px),"'} : 
                 'left: 123,');
$("#elem").css(JSON.parse(myObject + 'sexyness : 80'))

edited May 7, 2013 at 18:30

answered May 7, 2013 at 17:18

TonyH

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