Prevent combination of multiple selectors - sass

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.

Related

How to stop #mixin in SCSS from duplicating CSS?

I want to use one CSS style for two classes with mixin, but when I use mixin the final result will be 2 classes with the same CSS.
I have shared my code example below:
#mixin btnhover {
background-color: $bg-cl-blc;
color: $txt-cl-ff;
}
.btn-base {
font-size: 15px;
&:hover {
#include btnhover;
}
}
.btn-otln {
font-size: 15px;
&:hover {
#include btnhover;
}
}
**OUTPUT CSS**
.btn-base:hover {
background-color: #000;
color: #fff;
}
.btn-otln:hover {
background-color: #000;
color: #fff;
}
This is how Sass works - it allows for better organisation of the code, but this code is then compiled, retaining functionality and not caring about other aspects.
If you really care about how the output code is structured, I would suggest to create a separate style for the classes with the hover effect:
#mixin btnhover {
background-color: #000;
color: #fff;
}
.btn-base {
font-size: 15px;
}
.btn-otln {
font-size: 15px;
}
.btn-base:hover,
.btn-otln:hover {
#include btnhover;
}
But in this approach, the use of mixin (and Sass) is questionable (in this exact case).
Generally, when you use Sass (or any other compiled language), you don't really care about the output CSS.
This won't be your answer, but I want to show you another way to make a mixin
#mixin btnhover($back, $color) {
background: $back;
color: $color;
}
When you use it, you can plug in the values
#include mixin btnhover($bg-cl-blc, $txt-cl-ff)
That way you can use the mixin over and over in different places with different values
Just discovered this recently myself, it's a concept called 'placeholders' in SASS syntax (see example below). I've done my best to apply it to your situation below....
Put this in your .scss file:
$bg-cl-blc: #ff211a;
$txt-cl-ff: #fff;
$btn-base-size: 15px;
%btnhover {
background-color: $bg-cl-blc;
color: $txt-cl-ff;
}
%btn-common {
font-size: $btn-base-size;
}
.btn-base {
#extend %btn-common;
&:hover {
#extend %btnhover;
}
}
.btn-otln {
#extend %btn-common;
&:hover {
#extend %btnhover;
}
}
CSS output will look like this
.btn-otln:hover, .btn-base:hover {
background-color: #ff211a;
color: #fff;
}
.btn-otln, .btn-base {
font-size: 15px;
}
Great article written up on this here:
https://dev.to/kemotiadev/are-sass-mixins-really-that-lightweight-and-what-are-placeholders-119i

SCSS use parent style for children (texts)

I have next scss code:
.btn {
font-size: 24pt;
color: red;
&-small {
font-size: 17pt;
}
&-meddium {
font-size: 24pt;
}
&-big {
font-size: 36pt;
}
}
and i want that all my buttons (btn, btn-small, btn-meddium, btn-big) will have color RED.
P.S. I want realize it by only one class
Since you're using SCSS, you can use #mixin and #include:
#mixin btn {
color: red;
}
.btn {
#include btn;
}
.btn-smaall {
#include btn;
font-size: 17pt;
}
And to to do that for the rest of the button classes you want to create.
Another way would be to addjust your selectors as such:
.btn {
font-size: 24pt;
color: red;
&.btn-small {
font-size: 17pt;
}
&.btn-meddium {
font-size: 24pt;
}
&.btn-big {
font-size: 36pt;
}
}
And then you can create a small button by adding the class btn and btn-small to a button. e.g. <button class="btn btn-small>Test</button>
You can define classes like this (my preference): btn btn__small
Then you can trigger it with scss like you wanted:
.btn {
font-size: 24pt;
color: red;
&__small {
font-size: 17pt;
}
}

how to simply style in scss so that multiple elements will use same style

I am new to sass.
Here is my code in scss. Just wondering if this can be simplified further i.e i dont want to repeat the style color, text-decoration and transition.
a {
color: inherit;
text-decoration: none;
transition: all 0.3s;
}
div.menu-item-click {
&:hover, &:focus {
color: inherit;
text-decoration: none;
transition: all 0.3s;
}
}
Note exactly that use case is covvered better by ReSedano.
You cand do it using mixins:
#mixin mixinName {
color: inherit;
text-decoration: none;
transition: all 0.3s;
}
a {
#include mixinName;
}
div.menu-item-click {
&:hover, &:focus {
#include mixinName;
}
}
Also here is example with variables:
#mixin icon($width) {
width: $width;
stroke: currentColor;
}
.icon {
#include icon(25px);
}
And here is example with body
#mixin desktop ($xl: null) { // look here is default Value!
#media (min-width: if($xl, $xl, $screen-desktop)) {
#content; // here is true magic
}
}
.page {
#include desktop { // you may ignore variable because we have default
padding: 30px;
}
}
For this, maybe it is better using a placeholder with #extend directive (the output is less verbose than using a mixin):
%my-class {
color: inherit;
text-decoration: none;
transition: all 0.3s;
}
a {
#extend %my-class;
}
div.menu-item-click {
&:hover, &:focus {
#extend %my-class;
}
}
The output is:
a, div.menu-item-click:hover, div.menu-item-click:focus {
color: inherit;
text-decoration: none;
transition: all 0.3s;
}

How to extract common scss code from multiple files?

After converting a lot of redundant crappy css files into scss files, I have a bunch of scss files. I'm pretty sure there is a lot of common css repeated among these files and I would like to extract this code.
As an example, let's say I have this block of scss code (let's call it block A) :
.test {
color: white;
.toto {
background: red;
font-size: 12px;
}
}
And another block (that we'll call block B) :
.test {
color: black;
.toto {
background: blue;
font-size: 12px;
text-align: center;
}
}
I want to be able to extract the following common scss code from block A and B :
.test {
.toto {
font-size: 12px;
}
}
It seems like a simple task to do, but with a large list of long scss files, it's really painful to do it manually. After searching for a while I didn't find any tool for that.
An intermediary solution could be to convert sass code to a multi-dimensionnal associative array and to process arrays to find intersections, but I could not find any simple solution to do that either, so any help would be appreciated.
There are a few approaches but in this instance, I would opt for a variable:
$base-font-size: 12px;
.test {
color: white;
.toto {
background: red;
font-size: $base-font-size;
}
}
.test {
color: black;
.toto {
background: blue;
font-size: $base-font-size;
text-align: center;
}
}
Or you could add a toto mixin with some defaults and use that:
#mixin toto($background: red, $text-align: left, $font-size: 12px) {
.toto {
background: $background;
text-align: $text-align;
font-size: $font-size;
}
}
.test {
color: white;
#include toto();
}
.test {
color: black;
#include toto(blue, center);
}
EDIT: or use extend:
.font-size-12 {
font-size: 12px;
}
.test {
color: white;
.toto {
#extend .font-size-12;
background: red;
}
}
.test {
color: black;
.toto {
#extend .font-size-12;
background: blue;
text-align: center;
}
}

Using SASS to extend selectors with elements

I'm working on an SCSS stylesheet, and I have a rule that looks something like this:
.footer-link-row {
color: red;
ul& {
padding: 0;
}
}
I want the ul& line compile to the selector ul.footer-link-row. However, this selector returns a compiler error, and using a &ul compiles to .footer-link-row ul. What's the correct way to select something like this?
--Added--
To clarify, the eventual CSS I want out of this is:
.footer-link-row {
color: red;
}
ul.footer-link-row {
padding: 0;
}
You want something like the following:
ul {
padding: 0;
.footer-link-row {
color: red;
}
}
The ampersand is used to require that both selectors match
a { text-decoration: none;
&:hover { border-width: 1px }
}
// compiles to
a {
text-decoration: none;
}
a:hover {
border-width: 1px;
}
If you want the ul.footer-link-row try
ul {
&.footer-link-row {
padding: 0;
}
.footer-link-row {
color: red;
}
}
Your clarification indicates that you need two scopes.
ul {
&.footer-link-row {
padding: 0;
}
}
.footer-link-row {
color: red;
}

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