How to get in Javascript / jQuery only CSS override properties

Question

I am trying to obtain the list of CSS properties of an element that are overridden, example:

.div{
  position:absolute;
  top:10px;
}

When we use this code:

jQuery('.div').css('bottom');

I obtain:

bottom:-10px;

If I use get computed or jquery.css I will get the same result, but in fact I want to obtain:

empty or null

because I didn't specify / overridden.

Values expected in this example: top: 10px bottom: auto or null or nothing

https://codepen.io/anon/pen/dvvdVv

Again, I only wanted the properties that were overridden.

===================== EDIT ======================

My objective is only to get the value top,bottom from the following element:

<div id='box' class='giveItSomeTop' style="top:10px"></div>

and the css for the element, notice that I have only specified top:

.giveItSomeTop{
   top:10px;
}

Now I want to get the top value and the bottom value. If I do:

$('#box').css('top');

The result is: 10px but when I do:

$('#box').css('bottom');

I get 700px when I should get empty, nothing, null or ''

You can't.. When you query the `.css(x)` it is not trying to parse the css file... THAT is what you're looking for. If you take a look at the documenation, it says `Get the value of a computed style property` : https://api.jquery.com/css/ you will need to write a CSS file parser. — Pogrindis, Mar 10 '17 at 17:34
Thought so. That was exactly what we did. But could be a better way :( — Ignacio Correia, Mar 10 '17 at 18:46

score 1 · Answer 1 · answered Mar 10 '17 at 13:26

Its a really tricky thing you got there, my only suggestions here is:

Set the top / bottom with native javascript, so if you set top = 10px, you will get element.style.top == 10px, if you did not set this property, you will get "".

I added an example here, not the best solution but maybe it helps:

$( document ).ready(function() {
  var box = document.getElementById("box");
  
  // Set it like this
  var top = box.style.top = "20px";
  var bottom = box.style.bottom;
  
  // And you get 20px
  $('#top').text(top);
  
  // If unset you get ""
  $('#bottom').text(bottom);
  
  console.log(top);
});

#box {
  position: absolute;
  width:30px;
  height:30px;
  background-color:red;
  top:10px;
}

p {
  position:relative;
  top:40px;
  font-size:12px;
  color:black;
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id='box'></div>
<p id='top'></p>
<p id='bottom'></p>

I have edited the question. You answer does not work because we do not control how the CSS is setup. We have 5 css files that override the elements on this project. — Ignacio Correia, Mar 10 '17 at 17:31

score 1 · Accepted Answer · edited May 23 '17 at 12:25

Unfortunately you are not using the jQuery css() method how it is intended.

https://api.jquery.com/css/

Get the value of a computed style property for the first element in the set of matched elements or set one or more CSS properties for every matched element.

This means, it does not care what the CSS file has applied to this element, it will only look for what you have asked it to look for on the element.

What you will need to do it write a CSS file parser, and then query from that.. But it's another tool/library you will need to implement.

And there are answers for such a tool on StackOverflow : Parsing CSS in JavaScript / jQuery

And / Or : Convert CSS text to JavaScript object

From here you will be given a null|undefined for object properties which do not exist.

How to get in Javascript / jQuery only CSS override properties

2 Answers2