Using SASS variables to generate inline CSS

Question

I'm wanting to know if you can use SASS variables to generate inline CSS.

I'm aware you can do the following:

$my-class: yellow;
$yellow: #ffff00;

.#{$my-class} {
    color: $yellow;
}

And then inline the CSS using a Grunt task which outputs:

<div class="yellow" style="color: #ffff000;">Hello world</div> 

But is it possible to use a variable:

$font-family: Arial, sans-serif;

In such a way:

<div style="font-family: $font-family;">Hello world</div>

Which would output:

<div class="font-family: Arial, sans-serif;">Hello world</div>

I'm pretty sure you can't with basic SASS but it would good to hear some thoughts on it.

Answer 1

You can also use CSS variables

.box {
--difference: blue;
  background: red;
}

<div class="box">1</div>
<div class="box" style="background: var(--difference)">2</div>

Answer 2

You can use a html preprocessor such as jade together with a tool like sass-extract to accomplish this.

Here is an example where you would get the result in your example.

style.scss

$fontFamily: Arial;

template.jade

div(style="font-family: #{$fontFamily.value};") Hello world

index.js (could also be a grunt/gulp task)

const sassExtract = require('sass-extract');
const jade = require('jade');

const style = sassExtract.renderSync({ file: './style.scss' });
const html = jade.renderFile('./template.jade', style.vars.global);

console.log(html);

Outputs

<div style="font-family: Arial">Hello world</div>