jQuery's CSS property

Question

This may be a really stupid question, but I'm new to jQuery and am not understanding its css property like I thought I was,

It's blowing my mind that

-webkit-transform: rotate(30deg);

and

$("#cloud").css({
    "top" :  + 100 + "px",  
    "left" : "100px"});

work.

But that when I try and reuse that same idea here,

$("cloud").css({
    "-webkit-transform" : "rotate(30deg)"});

It doesn't work. It seems as though I am using the exact same syntax.

I apologize for the stupid question.

~Austin

Is that a typo or are you actually missing the # in your selector? — Ken Herbert
– Ken Herbert, Commented May 21, 2013 at 22:29
Funny how the smallest, most obvious mistakes can be the ones that make you tear out your hair in frustration. — Ian
– Ian, Commented May 21, 2013 at 22:47
Wow. Somebody should revoke my rights to publish. I spent an hour troubleshooting before I posted this up here. Maybe I should copy and paste next time. :P — aprohl5
– aprohl5, Commented May 21, 2013 at 23:01

Mircea · Accepted Answer · 2013-05-21 22:30:58Z

2

It actually works: http://jsfiddle.net/nzmfW/ but you messed up the selector:

$("cloud").css({"-webkit-transform" : "rotate(30deg)"});

should probably be :

$("#cloud").css({"-webkit-transform" : "rotate(30deg)"});

answered May 21, 2013 at 22:30

Mircea

11.7k26 gold badges67 silver badges98 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Mathijs Flietstra · Accepted Answer · 2013-05-21 22:30:48Z

1

Try

$("#cloud").css({"-webkit-transform" : "rotate(30deg)"});

You are missing a # before the id in the selector in the second line.

answered May 21, 2013 at 22:30

Mathijs Flietstra

13k3 gold badges42 silver badges73 bronze badges