Sass Mixin: Callback or Replace #content - sass

I don't know if Sass is able to do this, but it doesn't hurt to ask.
The Problem
Basically I have three colors pattern that are repeated in multiple sections of application, like blue, green and orange. Sometimes what changes is the background-color, or the border-color of the component... Sometimes is the text color of a child element, etc.
What I thought?
1. Replace a string pattern inside a content.
.my-class {
#include colorize {
background-color: _COLOR_;
.button {
border-color: _COLOR_;
color: _COLOR_;
}
}
}
2. Providing a callback variable for #content.
// This is just a concept, IT DOESN'T WORK.
#mixin colorize {
$colors: blue, green, orange;
#each $colors in $color {
// ...
#content($color); // <-- The Magic?!
// ...
}
}
// Usage
#include colorize {
background-color: $color;
}
I tried to implement such solutions, but without success.
Instead of it...
See below my workaround to get it partially working:
#mixin colorize($properties) {
$colors: blue, green, orange;
#for $index from 1 through length($colors) {
&:nth-child(#{length($colors)}n+#{$index}) {
#each $property in $properties {
#{$property}: #{nth($colors, $index)};
}
}
}
}
You can use this mixin that way:
.some-class {
#include colorize(background-color);
}
What will come output:
.some-class:nth-child(3n+1) {
background-color: blue;
}
.some-class:nth-child(3n+2) {
background-color: green;
}
.some-class:nth-child(3n+3) {
background-color: orange;
}
The problem? Well, I can't use it with child selectors.
Based on the above information, there is some magic solution for this case?

I think I figured out what you meant; it is a little (very) messy, but it should do what you want:
#mixin colorize($parentProperties,$childMaps) {
$colors: blue, green, orange;
#for $index from 1 through length($colors) {
&:#{nth($colors, $index)} {
#each $property in $parentProperties {
#{$property}: #{nth($colors, $index)};
}
}
#each $mapped in $childMaps {
$elem: nth($mapped,1);
$properties: nth($mapped,2);
#{$elem}:nth-child(#{length($colors)}n+#{$index}) {
#each $property in $properties {
#{$property}: #{nth($colors, $index)};
}
}
}
}
}
It would turn out to be:
/* -------------- USAGE ------------------*/
.some-class {
#include colorize(
background-color,( //Parent properties
(button, background-color), //Child, (properties)
(span, (background-color,border-color)) //Child, (properties)
)
);
}
/* --------------- OUTPUT ----------------*/
.some-class:nth-child(3n+1) {
background-color: blue;
}
.some-class button:nth-child(3n+1) {
background-color: blue;
}
.some-class span:nth-child(3n+1) {
background-color: blue;
border-color: blue;
}
.some-class:nth-child(3n+2) {
background-color: green;
}
.some-class button:nth-child(3n+2) {
background-color: green;
}
.some-class span:nth-child(3n+2) {
background-color: green;
border-color: green;
}
.some-class:nth-child(3n+3) {
background-color: orange;
}
.some-class button:nth-child(3n+3) {
background-color: orange;
}
.some-class span:nth-child(3n+3) {
background-color: orange;
border-color: orange;
}
Hope that that is what you are looking for :)

Related

One shared selector between multiple classes

How to apply the same focus state to multiple different classes?
Problem:
.btn {
&.error {
border-color: red;
}
&.primary {
border-color: green;
}
&:focus {
border-color: blue;
// this is not applied but i don't want to
// declare the same style to both classes
}
}
I understand this would be one option, but it is also not the prettiest option as i need to list them separately here
.btn {
&.error {
border-color: red;
}
&.primary {
border-color: green;
}
&.primary:focus, &.error:focus {
border-color: blue;
}
}
Are there any better ways?
Using & again in the nested rule is a good way for your purpose.
.btn {
&.error {
border-color: red;
}
&.primary {
border-color: green;
}
&.primary, &.error {
&:focus{
border-color: blue;
}
}
}

SCSS: How to use variables to write style classes ?

I was wondering whether there is a certain way to use variables that affect the style in SCSS.
I'm looking for something like:
var x = 1
.class1 {
if (x==1) {
background-color: red;
} else {
background-color: blue;
}
}
.class2 {
if (x==1) {
background-color: blue;
} else {
background-color: red;
}
}
You can use #if and #else
$x:1;
.class1 {
#if $x == 1 {
background-color: red;
} #else {
background-color: blue;
}
}
.class2 {
#if $x == 1 {
background-color: blue;
} #else {
background-color: red;
}
}

SASS Mixin: only apply hover style if class is on a link?

I'm attempting to set a bunch of background colours using a mixin. I'd also like to apply hover styling to these background colours IF the classes are assigned to a link element:
#mixin bg-color($color) {
background-color: $color;
&[ifthisisalink] {
&:hover {
background-color: darken($color, 10%);
}
}
}
.bg-blue {
#include bg-color(blue);
}
So if we have .bg-blue on a plain div, there is no hover color. But if .bg-blue is on a link, there is a hover color:
<div class="bg-blue">Hover on me and nothing happens.</div>
Hover on me and I go darker.
Is this possible in SASS?
You need #at-root:
#mixin bg-color($color) {
background-color: $color;
#at-root {
a#{&} {
&:hover {
background-color: darken($color, 10%);
}
}
}
}
.bg-blue {
#include bg-color(blue);
}

Prevent combination of multiple selectors

I'm trying to group all my vendor-specific stuff into a placeholder selector like this:
%search-bar-placeholder {
color: red;
}
.search-bar::-webkit-input-placeholder {
#extend %search-bar-placeholder;
}
.search-bar:-moz-placeholder {
#extend %search-bar-placeholder;
}
.search-bar::-moz-placeholder {
#extend %search-bar-placeholder;
}
.search-bar:-ms-input-placeholder {
#extend %search-bar-placeholder;
}
And then it compiles to this:
.search-bar::-webkit-input-placeholder, .search-bar:-moz-placeholder, .search-bar::-moz-placeholder, .search-bar:-ms-input-placeholder {
color: red; }
How can I make sure Sass doesn't put all the selectors together ? Like this:
.search-bar::-webkit-input-placeholder {
color: red;
}
.search-bar:-moz-placeholder {
color: red;
}
.search-bar::-moz-placeholder {
color: red;
}
.search-bar:-ms-input-placeholder {
color: red;
}
When looking at Extend/Inheritance at sass-lang.com it seems that the selectors will always be comma separated. Even if you add another property, it will keep the shared properties in the comma separated list, and add another selector just for that overridden value.
The way I achieved what you want is by using a mixin. Though it's not really the purpose of a mixin, it does get the job done. Your style is still centralized and you can print it out in each selector using a one liner too.
#mixin placeholder-properties() {
color: red;
font-weight: bold;
}
.search-bar::-webkit-input-placeholder {
#include placeholder-properties();
}
.search-bar:-moz-placeholder {
#include placeholder-properties();
}
.search-bar::-moz-placeholder {
#include placeholder-properties();
}
.search-bar:-ms-input-placeholder {
#include placeholder-properties();
}
The result will the following.
.search-bar::-webkit-input-placeholder {
color: red;
font-weight: bold;
}
.search-bar:-moz-placeholder {
color: red;
font-weight: bold;
}
.search-bar::-moz-placeholder {
color: red;
font-weight: bold;
}
.search-bar:-ms-input-placeholder {
color: red;
font-weight: bold;
}
Here's a fiddle.

Sass interpolate a variable name to string

We were provided a number of colors with specific hover-state colors associated:
$red: #cb333b;
$red-hover: #fe666e;
$brown: #544742;
$brown-hover: #877a75;
etc.
Since all the colors are formatted the same way, so I was hoping to write a mixin that takes the color's variable name, then concatenates -hover to the end. This is my first try:
#mixin button_colorizor($color) {
border-color: $color;
color: $color;
&:hover {
color: #{$color}-hover;
border-color: #{$color}-hover;
}
}
But what this does is output a color like this: #f1735f-hover. The same thing when I do this: color: #{$color+-hover};
You can create map of colors. And get color values by its names.
Demo on sassmeister.
$colors: (
red: #cb333b,
red-hover: #fe666e,
brown: #544742,
brown-hover: #877a75
);
#mixin button_colorizor($color) {
color: map-get($colors, $color);
border-color: map-get($colors, $color);
&:hover {
color: map-get($colors, $color + '-hover');
border-color: map-get($colors, $color + '-hover');
}
}
a {
#include button_colorizor(red);
}
span {
#include button_colorizor(brown);
}
This code is compiled to css:
a {
color: #cb333b;
border-color: #cb333b;
}
a:hover {
color: #fe666e;
border-color: #fe666e;
}
span {
color: #544742;
border-color: #544742;
}
span:hover {
color: #877a75;
border-color: #877a75;
}

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