I need to see if there is a way to parse out variables that are effectively, dynamically generated.
I have a list of 'color' definitions from a variables.scss file in my application:
$colorsList: "primary" "success" "info" "warning" "danger" "inverse";
I then have a number of variables that map to various colors:
$primary-color: #558bc3 !default;
$success-color: #259d2c !default;
$info-color: rgba(43, 169, 251, 0.39) !default;
$warning-color: #FAAA00 !default;
$danger-color: #c82e2d !default;
$inverse-color: #000;
I am trying to create an SCSS page that will effectively create some generated color blocks/css rules:
#each $color in $colorsList {
$col: "$color-#{$color}";
$bgcol: "$color-#{$color}";
.bg-#{$color} {
background-color: "#{$bgcol}";
}
.color-#{$color} {
color: "#{$col}";
}
}
This results in the following output:
.bg-primary {
background-color: "$color-primary"; }
.color-primary {
color: "$color-primary"; }
.bg-success {
background-color: "$color-success"; }
Essentially, what I would like to do - is have those variables in the output parsed to the variable values: $color-success gets changed with #259d2c, and so forth.
Is there a way to do this? Or maybe some workaround?
No, what you are trying is not possible, at least so for. A SCSS variable will be interpreted only once, as you can read on the official documentation:
Sass variables are simple: you assign a value to a name that begins with $, and then you can refer to that name instead of the value itself
So when you call background-color: "#{$bgcol}", $bgcol gets replaced by its value, which is just a string in the eye of SCSS, as any interpolated result:
Interpolation can be used in SassScript to inject SassScript into unquoted strings. This is particularly useful when dynamically generating names (for example, for animations), or when using slash-separated values. Note that interpolation in SassScript always returns an unquoted string.
What you can do that's common in the SCSS world is to map your color names to your colors with the help of a map, like so:
#use "sass:map";
$colorsList: "primary" "success" "info" "warning" "danger" "inverse";
$mapedColorsList: (
"primary": #558bc3,
"success": #259d2c,
"info": rgb(43 169 251 / 39%),
"warning": #faaa00,
"danger": #c82e2d,
"inverse": #000,
);
#each $color in $colorsList {
.bg-#{$color} {
background-color: map.get($mapedColorsList, $color);
}
.color-#{$color} {
color: map.get($mapedColorsList, $color);
}
}
Related
I want to export "$var" like variables defined as "--var" variables from a Scss file to be used in other files with some scss functions such as "darken()", but it says its not a color.
I'm coding a Vue/Vite website and implementing a theme-change button, that will write data at the HTML tag, and use it to define the theme.
<html lang="en" data-theme="dark"> ... </html>
The base.scss looks like this:
[data-theme="dark"] {
--primary: #c6c7df;
--secondary: #c95d63;
--text-light: #e2ebf5;
}
[data-theme="light"] {
--primary: #26298d;
--secondary: #c95d63;
--text-light: #2c3e50;
}
$primary: var(--primary);
$secondary: var(--secondary);
$text-light: var(--text-light);
I can use the "$var"-like variable at the end of the file, but when I use it inside a "darken($primary, 10%)" function, it gives the following error:
Error: $color: var(--primary) is not a color.
I know it's because I defined it as a CSS variable but if I define the variable inside the "[theme] {}" thing, it doesn't get innitialized and throws the error "Undefined variable"
I know it's a niche question, and I could use something like this, but I wish I could just export it as a scss variable.
Thank you very much.
Of course we cannot make a CSS variable lighten or darken. Think about, What will happen to the Name of CSS variable in the output?
So we have to define all CSS variables and assign desired colors, Before using them.
Solution
For this purpose, you can automatically create the list of CSS variables by #each and create a manual function by #function for easy use. This way gives us a standard css output and the best experience in Sass coding.
// Theme defines area:
$theme-colors:(
"primary": #940fe0,
"secondary": #e49f38,
"warning": #ffe600,
"info": #165ad6
);
$darken-percent: 20%;
$lighten-percent: 20%;
// Functions area:
:root {
#each $name, $color in $theme-colors {
$lighten: lighten($color, $lighten-percent);
$darken: darken($color, $darken-percent);
--color-#{$name}: #{$color};
--color-#{$name}-light: #{$lighten};
--color-#{$name}-dark: #{$darken};
}
}
#function cl($name, $color-tone) {
#if not map-get($theme-colors, $name+''){
#error "`#{$name}` is not defined in Colors Theme";
}
#if $color-tone == d {
#return var(--color-#{$name}-dark);
}
#if $color-tone == l {
#return var(--color-#{$name}-light);
}
#else{
#return var(--color-#{$name});
}
}
// Our custom function usage:
// Use 'b' to return Base color. Ex: cl(primary,b)
// Use 'l' to return Lighten color. Ex: cl(primary,l)
// Use 'd' to return Darken color. Ex: cl(primary,d)
.test {
color: cl(primary,b);
color: cl(warning,l);
}
Compiled css result:
:root {
--color-primary: #940fe0;
--color-primary-light: #bf61f4;
--color-primary-dark: #550980;
--color-secondary: #e49f38;
--color-secondary-light: #f0ca92;
--color-secondary-dark: #a06916;
--color-warning: #ffe600;
--color-warning-light: #fff066;
--color-warning-dark: #998a00;
--color-info: #165ad6;
--color-info-light: #6395ef;
--color-info-dark: #0c337a;
}
.test {
color: var(--color-primary);
color: var(--color-warning-light);
I tried to change the variable used in another variable from the mixin, but I couldn't change the variable value.
$color: white !default;
$var: "this #{$color} is a test" ;
#mixin test($value, $color) {
// here how to override the $color value
color: $value;
}
.a {
#include test($var, #ccc);
}
By default the $color value is in white, and this $color variable is used in another variable $var.
When I pass the $var inside a mixin to change its variable ($color) value, it is not changed. Anyone please suggest the solution for this?
Expected Output:
.a {
color: "this #ccc is a test";
}
Thanks in advance.
When you pass the $var variable as the argument, its value is not updated because it has already been assigned. Passing a new $color variable in the mixin won't update the variable. A workaround can be by using the set-nth function to replace the white with the color passed in the mixin:
$color: white !default;
$var: #{$color}, red;
#mixin test($value, $colour) {
// here how to override the $color value
$var: set-nth($value, 1, $colour);
color: $var;
}
.a {
#include test($var, #ccc);
}
TL;DR: How can you use SCSS to have CSS variables with a fallback for older browsers.
I'm trying to make sense of this article. In my opinion, you have to already be an advanced SASS user to understand it, which I'm not. To make matters worse, it's the only article I found on the subject.
Here is what I'm trying to achieve:
My scss should be along the lines of :
body {
#include v(background-color, primary)
}
then the processed CSS should be
body{
background: yellow; /* Yellow being defined above as the primary color */
background: var(--color-primary);
}
By playing around a bit, I can already get the value of the CSS variable like so:
$colors: (
primary: yellow,
);
:root {
#each $name, $color in $colors {
--color-#{$name}: $color;
}
}
#mixin background-color($color_) {
background: var(--color-#{$color_});
}
To use it:
body{
#include background-color(primary);
}
Which will result in this:
body {
background: var(--color-primary);
/* But the fallback is missing :(, I tried things with the map-get but it's really eluding me... */
}
If you're using Sass, you can automate fallbacks through a Sass mixin. Create a map of your CSS variable names and their values, and then you can look up those values in a mixin that outputs the fallback style and the preferred one
$vars: (
primary: yellow,
);
:root {
--primary: map-get($vars, primary);
}
#mixin var($property, $varName) {
#{$property}: map-get($vars, $varName);
#{$property}: var(--#{$varName});
}
The above mixin is used like so:
body {
#include var(background-color, primary);
}
and outputs the following CSS:
:root {
--primary: yellow;
}
body {
background-color: yellow;
background-color: var(--primary);
}
Et voilĂ :)
Update: Postcss Custom properties can do fallback and is way easier than the below code
step 1: declare scss variables
So first of all we want to put some variables in a $map, I'll go with color variables:
$colors: (
primary: #FFBB00,
secondary: #0969A2
);
step 2: automate css 4 var generation
// ripped CSS4 vars out of color map
:root {
// each item in color map
#each $key, $value in $colors {
--colors-#{$key}: $value;
}
}
What happens in root is : for each key and value in the colors map, we print the followng :
--colors-#{$key}: $value;
Which corresponds to css variable declarations. I believe the weird bit with #{} around the key is to not have spaces around the value.
Thus the result is:
--colors-primary: #FFBB00,
--colors-secondary: #0969A2
Note that the prefix (--colors-) is the same name as the scss color map above it. The why will become clear in last step.
step 3: Plenty of maps !
$props: (
background-color: $colors
);
$map-maps: (
background-color: colors
);
Here we add the map $props which maps a css property to the map containing the values. background-color will hold color, so the correct map is $colors.
map-maps is a copy of props where instead of the map we have the name of said map. (this is relative to the note in step 2).
Step 4 : let's make it work !
#mixin v($prop, $var) {
// get the map from map name
$map: map-get($props, $prop);
// fallback value, grab the variable's value from the map
$var-fall: map-get($map, $var);
// our css4 variable output
$var-output: var(--#{$map}-#{$var});
#{$prop}: $var-fall;
// css4 variable output
#{$prop}: $var-output;
}
body{
#include v(background-color, primary);
}
I simplified the code in the article quite a bit, it still works, for this example at least, the code in the article takes more into account.
Anyhow, here is what happens.
First, we call the mixin with:
#include v(background-color, primary);
Then upon entering,
$map: map-get($props, $prop); // map-get($props, background-color)
we have a variable called $map to which we assign the value that is inside the $props map at the key background-color which happen to be the $colors map. It's a bit of a maze but it's not that complicated once you resolve it.
Then for the fallback:
$var-fall: map-get($map, $var);
This simply gets the value of the map we just got (which is $colors) at the $var key (which happens to be primary). Thus the result is #FFBB00.
For the css var
$map-name: map-get($map-maps, $prop);
$var-output: var(--#{$map-name}-#{$var});
we recreate what we did to generate the var in the #each loop
Whole code would be :
$colors: (
primary: #FFBB00,
secondary: #0969A2
);
// ripped CSS4 vars out of color map
:root {
// each item in color map
#each $name, $color in $colors {
--colors-#{$name}: $color;
}
}
$props: (
background-color: $colors,
color: $colors
);
$map-maps: (
background-color: colors
);
#mixin v($prop, $var) {
// get the map from map name
$map: map-get($props, $prop);
// fallback value, grab the variable's value from the map
$var-fall: map-get($map, $var);
// our css4 variable output
$map-name: map-get($map-maps, $prop);
$var-output: var(--#{$map-name}-#{$var});
#{$prop}: $var-fall;
// css4 variable output
#{$prop}: $var-output;
}
body{
#include v(background-color, primary);
}
Now this is a simplification of what is done in the article. You should check it out to have code a bit more robust.
I assume you are aware of the reason why it didn't show the fallback. But since it's an answer I will explain the reasons
The current mixin block has only one background property which makes the sass compiler to generate only one property. I don't think sass can identify whether 'var' is supported in browser or not. So, we have to explicitly specify if we need the fallback.
Since you already have the map all you need is to get the value by giving the key 'primary'
#mixin background-color($color_) {
background: var(--color-#{$color_});
background: map-get($colors, primary);
}
This will add the background: yellow to the body class always. Alternatively if you want to control the addition of the fallback based on condition. You can do like this
#mixin background-color($color_, $showFall) {
background: var(--color-#{$color_});
#if $showFall {
background: map-get($colors, primary);
}
}
and call like this
body{
#include background-color(primary, true);
}
Code pen for the same
https://codepen.io/srajagop/pen/xdovON
Note: I am writing the answer under the assumption that you want only the background-color to work and not all the other properties like mentioned in that post. For that you need to create a proper data structure
Google Materialize defines their color variables in this file like so:
$red: (
"lighten-5": #FFEBEE,
"lighten-4": #FFCDD2,
"lighten-3": #EF9A9A,
"lighten-2": #E57373,
"lighten-1": #EF5350,
"base": #F44336,
"darken-1": #E53935,
"darken-2": #D32F2F,
"darken-3": #C62828,
"darken-4": #B71C1C,
"accent-1": #FF8A80,
"accent-2": #FF5252,
"accent-3": #FF1744,
"accent-4": #D50000
);
I would like to use the colors with something like this:
.light-red-border {
border: 1px solid $red-lighten-1;
}
How do I call those variables directly? They're used in other I can't find anything on this syntax mentioned anywhere.
Edit: I looked around a bit more in the Materialize Github and found some examples in _variables.scss:
$primary-color: color("materialize-red", "lighten-2") !default;
$primary-color-light: lighten($primary-color, 15%) !default;
$primary-color-dark: darken($primary-color, 15%) !default;
However, if you can direct me to an article discussing the syntax used to define the variables I would greatly appreciate it.
You're looking for map-get($red, 'lighten-5'); if you simply want the corresponding value from this map.
To expand on #weirdpanda's answer, these SASS maps need to be iterated upon, which then compiles into more CSS rules.
In the file you linked to, each of those colors is placed in a larger map of $colors:
$colors: (
"materialize-red": $materialize-red,
"red": $red,
"pink": $pink,
"purple": $purple
//...
);
This colors map is then iterated upon, producing matching class selectors like red.lighten-5 in the CSS.
#each $color_name, $color in $colors {
#each $color_type, $color_value in $color {
#if $color_type == "base" {
.#{$color_name} {
background-color: $color_value !important;
}
.#{$color_name}-text {
color: $color_value !important;
}
}
#else {
.#{$color_name}.#{$color_type} {
background-color: $color_value !important;
}
.#{$color_name}-text.text-#{$color_type} {
color: $color_value !important;
}
}
}
}
To actually use these variables within SASS files (for instance, after importing _color.scss, you can access a map value with map-get($map, $key). For instance:
.my-class{
color: map-get($red, 'lighten-5');
}
Edit: regarding the color function: The $primary-color: color("materialize-red", "lighten-2") line you posted is using a function defined in _color.scss, which ostensibly does the same thing as map-get but checks if the key exists in the map with map-has-key and raises a warning if it cannot be found.
#function color($color, $type) {
#if map-has-key($colors, $color) {
$curr_color: map-get($colors, $color);
#if map-has-key($curr_color, $type) {
#return map-get($curr_color, $type);
}
}
#warn "Unknown `#{name}` in $colors.";
#return null;
}
This syntax notation is called the SASS Maps notation and it is the SASS-equiv of a hash-map. Read more about it here.
This is primarily a syntax question, I'm pretty sure I can do this but I don't know the best way to unquote, declare variables, and concatenate strings in the for loop. Essentially though I want to take this map I made and generate the variables for the colors as well as some classes that will assign styles such as font-color and background-color.
$color-collection: ('white': '#FFFFFF', 'black': '#000000', 'goldenrod': '#F59600');
#for $i from 0 to length($color-collection) {
//create a variable for each color name, e.g. $white: #FFFFFF;
//create a class for each variable name with a style, e.g. .bg-color-white {background-color: $white;}
}
First you need to use #each instead of #for.
Second you can just get the two variables you need in the loop itself. The first variable will be the key, and the second the value.
Third we can use sass interpolation to include the variable in a class name and to unqote the value in a selector. I think you could also use unqote($color-value) for this, or you could just write your color values in the array without quotation-marks.
Se example below.
$color-collection: ('white': '#FFFFFF', 'black': '#000000', 'goldenrod': '#F59600');
#each $color-name, $color-value in $color-collection {
.bg-color-#{$color-name} {
background-color: #{$color-value};
}
//create a variable for each color name, e.g. $white: #FFFFFF;
//create a class for each variable name with a style, e.g. .bg-color-white {background-color: $white;}
}
http://sassmeister.com/gist/e321b7e36ed0e74cd2ce
Since you are using libsass here's some code that would work with older Sass syntax:
$color-collection: ('white' '#FFFFFF'), ('black' '#000000'), ('goldenrod' '#F59600');
#each $color in $color-collection {
$color-name: nth($color, 1);
$color-value: nth($color, 2);
.bg-color-#{$color-name} {
background-color: #{$color-value};
}
}
http://sassmeister.com/gist/28caaf9d5e644d5db0cc