Jquery function css [duplicate]

Question

I want know if this script is bad, not working for me :

jQuery(".template_content_img").css({
    width: "50%",
    height: "150px",
    //// Here gives me error 
    font-size: "12px",
    margin-bottom: "23px",
});

I don´t know if the problem it´s is "width:200px" or "width","150px"

These are my dudes, by another side I see also many people use animate for this and no use css but why do you suppose the animate create effect and transformation of divs and by this only I need change somethings but it's not working.

Regards !!!

@PSL - good but 50% or 150px is string so must be in quotes or apostrofs — Shaddow
– Shaddow, Commented Jun 20, 2013 at 2:22
@user2501504 Put the property keys in quotes. "font-size". — PSL
– PSL, Commented Jun 20, 2013 at 2:26

Shaddow · Accepted Answer · 2013-06-20 02:29:56Z

5

Check jQuery .css documentation

jQuery(".template_content_img").css({
    "width": "50%",
    "height": "150px"
});

jQuery(".template_img").css({
    "width": "70%",
    "height": "100px"
});

EDIT: If you want add font-size or margin-bottom, just use classic CSS attributes/values

jQuery(".template_content_img").css({
    "width": "50%",
    "height": "150px"
    "font-size": "12px",
    "margin-bottom": "23px",
});

edited Jun 20, 2013 at 2:29

answered Jun 20, 2013 at 2:15

Shaddow

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2 Comments

user2501504 Over a year ago

And one question how i can put this or use if i want put also for example othe things as fot-size , font-weight , margin-bottom , etc , thank´s for the help

Shaddow Over a year ago

just use classic CSS attributes/values

Barmar · Accepted Answer · 2013-06-20 02:42:10Z

1

For multi-word CSS attributes, either use quotes:

"font-size": "12px",

or camelCase:

fontSize: "12px",

The object has to be a valid Javascript object literal, and delimiters like - are not allowed in unquoted property names.

answered Jun 20, 2013 at 2:42

Barmar

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1 Comment

Niet the Dark Absol Over a year ago

+1 for having both options and the explanation.

Jorge · Accepted Answer · 2013-06-20 02:19:41Z

0

You have to parse the css properties as a javascript Object using literals:

$('.class').css({
   width : '50%',
   height : '150px'
});

edited Jun 20, 2013 at 2:19

Jorge

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answered Jun 20, 2013 at 2:16

Victor Castillo Torres

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2 Comments

Niet the Dark Absol Over a year ago

-1 irrelevant to the question

Niet the Dark Absol Over a year ago

Because it completely fails to address the cause of the error.

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