SCSS variable in background image with SVG image data URI

Question

In the following SCSS, I would like to use the fg-color variable within the url background-image attribute.

$fg-color: #ff6464;

i.icon-back {
  background-image: url("data:image/svg+xml;charset=utf-8,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 12 20'><path d='M10,0l2,2l-8,8l8,8l-2,2L0,10L10,0z' fill='%23ff6464'/></svg>");
}

At the moment the value, of the variable in simply repeated within the SVG string, like so:

fill='%23ff6464'

I would like this to be automatically updated whenever the fg-color variable is updated.

How can I do this?

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UPDATE:

This input SCSS:

i.icon-back {
  background-image: url("data:image/svg+xml;charset=utf-8,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 12 20'><path d='M10,0l2,2l-8,8l8,8l-2,2L0,10L10,0z' fill='$fg-color'/></svg>");
}

Outputs this CSS:

i.icon-back {
  background-image: url("data:image/svg+xml;charset=utf-8,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 12 20'><path d='M10,0l2,2l-8,8l8,8l-2,2L0,10L10,0z' fill='$fg-color'/></svg>");
}

... it is exactly the same - the variable is not processed.

---

NOTE:

I have reviewed these questions, that appear to be similar, but are not the same:

• Creating a Data URI's from SCSS
• Using an Data URI SVG as a CSS background image

Answer 1

First, create a SASS function. This (ab)uses string interpolation to convert the colour to a string, and then strips the first character (should always be '#'), and puts '%23' in its place (URL encoded form of '#').

@function url-friendly-colour($colour) {
    @return '%23' + str-slice('#{$colour}', 2, -1)
}

Then use the function - but in order for it to work within the string, it must be interpolated, using #{}

i.icon-back {
  background-image: url("data:image/svg+xml;charset=utf-8,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 12 20'><path d='M10,0l2,2l-8,8l8,8l-2,2L0,10L10,0z' fill='#{url-friendly-colour($fg-color)}'/></svg>");
}

The caveat with this approach, of course, is that it will not work with colours that are specified by colour name instead of their hexadecimal colour.

E.g. #f00 works but red will not.