How to add multiple CSS elements to a div using jQuery?

Question

I would like to do this:

<div style="float: left;
            width: 59px;
            background: transparent url('http://download.com/47.jpg') no-repeat scroll -132px -1px;"
     class="cIconSmall">
</div>

and I'm thinking I should use this:

$("#YourElementID").css({
    float: "left",
    width: "59px",
    background: "transparent url('http://download.com/47.jpg') no-repeat scroll -132px -1px"
});

Any ideas?

Thanks

The only thing wrong with the code you first posted is that instead of $('#YourElementId') , you needed to do $('.YourClassName') since the div has a Class and not an Id — Dutchie432
– Dutchie432, Commented Apr 19, 2011 at 18:24

hunter · Accepted Answer · 2011-04-19 18:20:36Z

48

You're thinking correctly. Using the css(map) method is the way to go.

$(".cIconSmall").css({
    float: "left", 
    width: "59px", 
    background: "transparent url('http://download.com/47.jpg') no-repeat scroll -132px -1px" 
});

http://api.jquery.com/css/

A map of property-value pairs to set.

Might be nicer as a css class, though... then you can just write $(".cIconSmall").addClass("icon47"); but there's a time for everything...

answered Apr 19, 2011 at 18:20

hunter

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3 Comments

Patrioticcow Over a year ago

thanks, i was thinking to add a class, but my link will be different every time.

Ahmed Mahmoud Over a year ago

I'm sorry but it doesnot work until i put float ,width,backroung around quotes

Andrew L Over a year ago

I had to do this for it to work. $('#selector').css({'property': 'val','property': 'val'});

VPK · Accepted Answer · 2017-02-28 12:35:20Z

2

$(".yourClass").css({
    float: "left",
    width: "59px",
    background: "transparent url('http://download.com/47.jpg') no-repeat scroll -132px -1px"
});

edited Feb 28, 2017 at 12:35

VPK

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answered Feb 28, 2017 at 11:22

Superman

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Comments

T Beatty · Accepted Answer · 2019-08-15 17:38:59Z

-1

I found that Andrew Lohr's comment in the accepted solution is for sure the only way I got this to work in my script: $('#selector').css({'property': 'val','property': 'val'}); (and thanks Andrew).

answered Aug 15, 2019 at 17:38

T Beatty

174 bronze badges

1 Comment

FabricioG Over a year ago

Comments should be added as a comment, not an answer

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