Tags: tinylesson

2

sparkline

Inlining SVG background images in CSS with custom properties

Here’s a tiny lesson that I picked up from Trys that I’d like to share with you…

I was working on some upcoming changes to the Clearleft site recently. One particular component needed some SVG background images. I decided I’d inline the SVGs in the CSS to avoid extra network requests. It’s pretty straightforward:

.myComponent {
    background-image: url('data:image/svg+xml;utf8,<svg> ... </svg>');
}

You can basically paste your SVG in there, although you need to a little bit of URL encoding: I found that converting # to %23 to was enough for my needs.

But here’s the thing. My component had some variations. One of the variations had multiple background images. There was a second background image in addition to the first. There’s no way in CSS to add an additional background image without writing a whole background-image declaration:

.myComponent--variant {
    background-image: url('data:image/svg+xml;utf8,<svg> ... </svg>'), url('data:image/svg+xml;utf8,<svg> ... </svg>');
}

So now I’ve got the same SVG source inlined in two places. That negates any performance benefits I was getting from inlining in the first place.

That’s where Trys comes in. He shared a nifty technique he uses in this exact situation: put the SVG source into a custom property!

:root {
    --firstSVG: url('data:image/svg+xml;utf8,<svg> ... </svg>');
    --secondSVG: url('data:image/svg+xml;utf8,<svg> ... </svg>');
}

Then you can reference those in your background-image declarations:

.myComponent {
    background-image: var(--firstSVG);
}
.myComponent--variant {
    background-image: var(--firstSVG), var(--secondSVG);
}

Brilliant! Not only does this remove any duplication of the SVG source, it also makes your CSS nice and readable: no more big blobs of SVG source code in the middle of your style sheet.

You might be wondering what will happen in older browsers that don’t support CSS custom properties (that would be Internet Explorer 11). Those browsers won’t get any background image. Which is fine. It’s a background image. Therefore it’s decoration. If it were an important image, it wouldn’t be in the background.

Progressive enhancement, innit?

A tiny lesson in query selection

We have a saying at Clearleft:

Everything is a tiny lesson.

I bet you learn something new every day, even if it’s something small. These small tips and techniques can easily get lost. They seem almost not worth sharing. But it’s the small stuff that takes the least effort to share, and often provides the most reward for someone else out there. Take for example, this great tip for getting assets out of Sketch that Cassie shared with me.

Cassie was working on a piece of JavaScript yesterday when we spotted a tiny lesson that tripped up both of us. The script was a fairly straightforward piece of DOM scripting. As a general rule, we do a sort of feature detection near the start of the script. Let’s say you’re using querySelector to get a reference to an element in the DOM:

var someElement = document.querySelector('.someClass');

Before going any further, check to make sure that the reference isn’t falsey (in other words, make sure that DOM node actually exists):

if (!someElement) return;

That will exit the script if there’s no element with a class of someClass on the page.

The situation that tripped us up was like this:

var myLinks = document.querySelectorAll('a.someClass');

if (!myLinks) return;

That should exit the script if there are no A elements with a class of someClass, right?

As it turns out, querySelectorAll is subtly different to querySelector. If you give querySelector a reference to non-existent element, it will return a value of null (I think). But querySelectorAll always returns an array (well, technically it’s a NodeList but same difference mostly). So if the selector you pass to querySelectorAll doesn’t match anything, it still returns an array, but the array is empty. That means instead of just testing for its existence, you need to test that it’s not empty by checking its length property:

if (!myLinks.length) return;

That’s a tiny lesson.