I have the following HTML <div class="parent green"></div>
The green class may or may not be added. It is dynamic. It may also be another name.
In SASS how do I give properties to a .child element of parent when class green is chained to it?
I tried:
.parent {
.child {
.green & {
color: green;
}
}
}
It doesn't work.
I also tried the following which works but I am looking for something similar to the sass above. The code will become repeatable below because I have to add child each time for every dynamic class.
.parent {
&.green {
.child {
color: green;
}
}
}
I'm trying to get a structure like this if possible with sass:
.parent {
.child {
.green & { /* when .parent.green */
color: green;
}
.blue & { /* when .parent.blue */
color: blue;
}
.text-align-right & { /* when .parent.text-align-right */
text-align: right;
}
etc...
}
}
& is treated as parent selector reference in Sass, because of this your code doesn't work since it refers wrong selector.
Use of & directly will not help here, but your goal can be achieved by using mixins, for example:
#mixin child($class) {
&.#{$class} {
.child {
#content;
}
}
}
.parent {
#include child(green) {
color: green;
}
#include child(blue) {
color: blue;
}
#include child(text-align-right) {
text-align: right;
}
}
This piece of code produces result that you want to get, you can check in by yourself on sassmeister.
Related
This is the usual list where one of the items is .open
for this I want to check if the parent (in this case is .item but its not relevant i think) has a specific class.
I've tried > but it doesnt seem to work.
Essentially how to put this:
&.open .info {
display: none;
}
&.open .inner-info {
display: flex;
}
inside of the their specific classes:
.info {
display: flex;
/* some other stuff */
}
.inner-info {
display: none;
/* some other stuff */
}
all of this is inside an .item{} block
So how do i have it so that i only have two blocks inside the .item{}?
It seems overkill to me, but you can use a hacky way to do that using a mixin and various functions. Please note that this will work for your specific example but probably not for something else.
I used the helper functions str-to-list and nth-delete, which are not native to SASS.
#mixin parentWithClass($class) {
$parent: nth-delete(str-to-list(#{&}), -1);
#at-root #{selector.replace(&, $parent, #{$parent}#{$class})} {
#content;
}
}
.item {
.inner {
color: blue;
#include parentWithClass(".open") {
color: orange;
}
}
.inner-info {
color: red;
#include parentWithClass(".open") {
color: grey;
}
}
}
You can also nest -info in inner.
In Sass/SCSS, is there any way to wrap an existing rule like this:
.foo {
color: red;
}
to produce output like this:
.foo,
.bar .foo {
color: red;
}
?
I know you can use the parent selector to accomplish this:
.foo {
&,
.bar & {
color: red;
}
}
but that requires modifying the existing rule. Ideally there'd be something magical like this that you could do at the root:
#noop,
.bar {
.foo {
color: red;
}
}
(I've tried #at-root but that doesn't work in list selectors, and just an ampersand & doesn't work at the root)
If you want to wrap a class without modifying it, you need to create a mixin which extends from that class.
The mixin I created here allows to pass a list as a parameter in order to put multiple selectors.
#mixin wrap($content){
#if (type-of($content) == string){
& #{$content} {
#extend #{$content};
}
}#else if (type-of($content) == list){
#each $class in $content{
& #{$class} {
#extend #{$class};
}
}
}
}
.red {
color: red;
}
.blue {
color: blue;
}
.wrap-1{
#include wrap('.red' '.blue');
}
.wrap-2{
#include wrap('.red');
}
Outputs:
.red, .wrap-2 .red, .wrap-1 .red {
color: red;
}
.blue, .wrap-1 .blue {
color: blue;
}
I'm trying to remove some duplication in my scss selector.
.container {
.operation {
color: green;
}
.row.is-active &,
.row:hover & {
.operation {
color: inherit;
}
}
}
I tried rewriting it like this:
.container {
.operation {
color: green;
}
.row & {
&.is-active, &:hover {
.operation {
color: inherit;
}
}
}
}
However, this causes .is-active to be applied to .container instead of .row
How can I target the syntactical parent when using the ampersand ?
I took some time to answer the question again, as I mis-understood it initially. Unfortunately there is absolutely no way possible to do this in SASS at the moment. Even when trying to make use of the more advanced SASS functions to manipulate selectors and strings it is not possible.
There is some Good News
It is possible to do using Stylus.
I have created a live Example on codepen.
// Stylus - not SASS
.container {
.operation {
color: green;
}
.row {
^[-1..-1]:is-active ^[0], ^[-1..-1]:hover ^[0] {
.operation {
color: inherit;
}
}
}
}
I hope this helps you in some way, at the very least it might provide you with an option, but unfortunately SASS cannot achieve what you are attempting.
I have the following sample code:
.level1 {
// css
.level2 {
// css
. level3 {
// css
color: red;
}
}
And then
.level1.blue .level .level3 {
color: blue
}
I would like to put the second rule somehow on the first bit of code, so that I don't repeat the structure again and I have both color possibilities above, is this possible in anyway?
I wasn't planning on answering my own question, but it seems that I found out exactly what I was looking for only it has recently being added to sass and will be available on sass 3.4. I believe there's a prerelease to tried but I havent tried it yet.
Basically what I was looking has been answered to me on github:
https://github.com/sass/sass/issues/286#issuecomment-49112243
So on 3.4 you can do:
.level1 {
.level2 {
.level3 {
#at-root #{selector-append(".blue", &)} {
color: blue;
}
color: red;
}
}
}
which is exactly what I was looking for.
There's a bunch of addition related to the parent selector (&), you can learn more from it at https://github.com/sass/sass/issues/1117
Bear in mind though, that at the time of writing this answer, all of this is rather new.
Also see: https://github.com/sass/sass/blob/master/doc-src/SASS_CHANGELOG.md
And: http://www.phase2technology.com/blog/everything-you-need-to-know-about-sass-3-4/
This:
#mixin level3color($color) {
.level2 {
.level3 {
color: $color;
}
}
}
.level1 {
#include level3color(#FF0000);
&.blue {
#include level3color(#0000FF);
}
}
produces this:
.level1 .level2 .level3 {
color: red;
}
.level1.blue .level2 .level3 {
color: blue;
}
Gotta love mixins!
EDIT:
This is still pretty clean (or at least clean considering what you're trying to do) because you can still have your structure there.
.level1 {
// css
.level2 {
// css
.level3 {
// css
color: red;
}
}
&.blue { #include level3color(blue); }
&.yellow { #include level3color(yellow); }
}
A simple example:
.child{
background-color:red;
.parent:hover &{
background-color:blue;
}
}
goes into
.child {
background-color: red;
}
.parent:hover .child {
background-color: blue;
}
http://sassmeister.com/gist/e994e056d3cc3b342e2c
I have the following sass code:
.class{
label{
color:#fff;
.disabled &{color:#333; }
}
}
which outputs
.disabled .class label
Is there a way to output the parent selector without any grandparent selectors being included? Like so:
.disabled label
There's no way I know of in SASS to pick and choose from ancestor selectors when using a parent reference. With your code, though, a little reorganization can get you the same result:
label {
.class & {
color: #fff;
}
.disabled & {
color:#333;
}
}
Compiles to:
.class label {
color: #fff; }
.disabled label {
color: #333; }
Even though hopper is not enterly wrong, you can actually select grand-parent with variables.
You can achieve what you want with this:
.class{
label{
color:#fff;
$selector: nth(&,1);
$direct-parent: nth($selector, length($selector));
#at-root #{$direct-parent} {
.disabled &{color:#333; }
};
}
}
Which will generate this css:
.class label {
color: #fff;
}
.disabled label {
color: #333;
}
The parent selector is always a reference to the entire resolved selector from the previous level of nesting. There is no concept of "parent" or "grandparent", especially when concatenating selectors or using the parent selector at the end muddies the water.
Disclaimer: I do not recommend doing this unless you really really need to.
Starting with Sass 3.4, you can extract portions of a selector by using & as a variable. When used this way, you'll get a list of list of strings (which can be looped over, etc.).
Extracting a part or slice of a selector
This function here uses the same style of arguments as the string-slice function:
#function selector-slice($sel, $start: 1, $end: -1) {
$collector: ();
#each $s in $sel {
// calculate our true start and end indices when given negative numbers
$_s: if($start > 0, $start, length($s) + $start + 1);
$_e: if($end > 0, $end, length($s) + $end + 1);
$c: ();
#for $i from $_s through $_e {
$c: append($c, nth($s, $i));
}
// prevent duplicates from creeping in
#if not index($collector, $c) {
$collector: append($collector, $c);
}
}
#return $collector;
}
/* complex example */
.one-a, .one-b {
two {
three {
color: red;
&:before {
#at-root #{selector-slice(&, 2, 3)} {
color: green;
}
}
}
}
}
/* your example */
.class {
label {
color:#fff;
#at-root #{selector-slice(&, -1, -1)} {
.disabled & {
color:#333;
}
}
}
}
Output:
/* complex example */
.one-a two three, .one-b two three {
color: red;
}
two three:before {
color: green;
}
/* your example */
.class label {
color: #fff;
}
.disabled label {
color: #333;
}
As an added bonus, you can use this function to reverse the order of the selectors by passing in the larger index before the smaller one.
.one-a, .one-b {
two {
three {
color: red;
&:before {
#at-root #{selector-slice(&, 3, 2)} {
color: green;
}
}
}
}
}
Output:
.one-a two three, .one-b two three {
color: red;
}
three:before two {
color: green;
}
Related: Modifying the middle of a selector in Sass (adding/removing classes, etc.)
Replacing one class with another
Alternately, you could just use the selector-replace function from the standard library if what you're looking to do is replace one class with another.
.class {
label {
color:#fff;
#at-root #{selector-replace(&, '.class', '.disabled')} {
color:#333;
}
}
}
Output:
.class label {
color: #fff;
}
.disabled label {
color: #333;
}