I want to get something like
.classA, .classB, .classC, .classD, .classE {
color: white;
}
.classA .classI, .classB .classI, classC .classI {
background: red;
}
is this possible with something like
.classA, .classB, .classC {
color: white;
.classI {
background: red;
}
}
class D and E shouldn't geht the class I.
hopefully you know, what I mean
For your case you shouldn't use .classD and .classE at the top level because the nested selector doesn't apply to all of them.
You should just use .classA, .classB, .classC and then use :extend for the other two.
.classA, .classB, .classC {
color: white;
.classI {
background: red;
}
}
.classD, .classE {
&:extend(.classA);
}
When compiled it would result in the following CSS:
.classA, .classB, .classC, .classD, .classE {
color: white;
}
.classA .classI, .classB .classI, .classC .classI {
background: red;
}
Yes like that:
.classA {
color: white;
&.classI {
background: red;
}
}
Related
How could I disable background-color in .button.search so it would fallback to $red value? I can't remove it; I can only overwrite it.
I have
.button {
background-color: {$red};
}
and
.button.search {
background-color: #000;
}
Don't need for any additional setting in search.
.button {
background-color: $red;
}
.button.search {
/* no background-color setting would fallback to $red*/
}
I would do it like this so you can extend the style from .search and it will always fallback with whatever you define and incase you want to have new value for the .active class you can just write background-color: green; after #extend .search;
.search {
background-color: red;
&.active {
#extend .search;
// background-color: green;
}
}
result will be like that
.search, .search.active {
background-color: red;
}
and if you will do that
.search {
background-color: red;
&.active {
#extend .search;
background-color: green;
}
}
and result will be like that
.search, .search.active {
background-color: red;
}
.search.active {
background-color: green;
}
Given this scss
.root {
color: red;
&-child {
color: blue;
small & {
font-size: 80%;
}
}
}
This is the CSS I get:
.root {
color: red;
}
.root-child {
color: blue;
}
small .root-child {
font-size: 80%;
}
I want to style .root-child on small differently so the rule I need is:
small.root-child {
font-size: 80%;
}
(Notice no whitespace after small)
How can I do that?
You need to use #at-root and that will remove the white space in your selector, as well as it will be a valid syntax so no issues while you try to compile.
.root {
color: red;
&-child {
color: blue;
#at-root small#{&} {
font-size: 80%;
}
}
}
You can use #at-root like this:
SCSS
.root {
color: red;
&-child {
color: blue;
#at-root {
small#{&} {
font-size: 80%;
}
}
}
}
Compiled:
.root {
color: red;
}
.root-child {
color: blue;
}
small.root-child {
font-size: 80%;
}
Just wanted to know if there exist any way to extend just first two of separated Class like in example, or either exist any other option like creating a specific Class
.background{background:red} and use it as extension instead of a separated Class (but i don't wanted to output in CSS a class .background).
EXAMPLES:
SASS:
.foo {
background:red
}
.foo {
color:red
}
.bar {
#extend .foo;
}
.foo {
font-size: 16px
}
LESS:
.foo {
background:red
}
.foo {
color:red
}
.bar {
&:extend(.foo);
}
.foo {
font-size: 16px
}
The output in CSS will be:
.foo, .bar {
background: red;
}
.foo, .bar {
color: red;
}
.foo, .bar {
font-size: 16px;
}
But I want to be like this:
.foo, .bar {
background: red;
}
.foo, .bar {
color: red;
}
// No .bar class here
.foo {
font-size: 16px;
}
What way should i follow to make this happened?
You've got your inheritance backwards. bar does not extend foo, foo extends bar:
LESS:
.bar {
background-color: red;
}
.bar {
color: red;
}
.foo {
&:extend(.bar);
font-size: 16px;
}
Produces
CSS:
.bar,
.foo {
background-color: red;
}
.bar,
.foo {
color: red;
}
.foo {
font-size: 16px;
}
I try to understand BEVM+SCSS philosophy.
I don't know how to extend V from BE in this case.
What I want to achieve:
.block {
&__element {
background-color: black;
&--variation-a {
#extend &__element; //won't work
color: red;
}
&--variation-b {
#extend &__element; //won't work
color: green;
}
}
}
What I want to avoid:
.block {
&__element {
background-color: black;
&--variation-a {
#extend .block__element; //work but ugly
color: red;
}
&--variation-b {
#extend .block__element; //work but ugly
color: green;
}
}
}
The only way I've found it's to have a kind of %element { ... } aside and extends from it, but it's not exactly what I want.
You can use variables. $b to store block name and $e to store element name.
Sassmeister demo.
.block {
$b: &;
&__element {
$e: #{$b}__element;
background-color: black;
&--variation-a {
#extend #{$e};
color: red;
}
&--variation-b {
#extend #{$e};
color: green;
}
}
}
But it's bad practice to nest element styles by modifier. Modifier must only override styles.
sorry but it's confusing to me, somebody knows how it's possible or it's not possible..
#divp {
background-color: lightgrey;
.odiv {
background-color: yellow;
.pp { background-color: black; }
a { color:red; }
}
.pp { background-color: lightgreen; }
a { color:blue; }
}
#divw {
background-color: lightblue;
.odiv {
background-color: blue;
.pp { background-color: white; }
a { color:yellow; }
}
.pp { background-color: green; }
a { color:lightblue; }
}
i want create divs with internal css rules and i dont want to write all the time the same..... like
#diw .odiv .pp { background-color: white }
#diw .odiv .a { color: white }
#diw .odiv .other { color: blue }
is it possible?
Nesting selectors is not possible, but you might want to checkout CSS preprocessors, which will let you do this. http://lesscss.org/ for example.
It is not possible in standard CSS. But it is possible in Sass (and other CSS Preprocessors): http://sass-lang.com/guide#3
It works exactly as you posted in your question:
.div1{
background-color: red;
p{ font-size: 18px; }
}
Will output this:
.div1{ background-color: red }
.div1 p{ font-size: 18px; }
Check out http://sassmeister.com/ for a way to play around with Sass.