I have global css class with the name of i-style. I want to change its style depends on selector. I know that it can be done by using class and &. selector. But I want to get it done by giving selector only.
for example
i-style{
border:solid 1px black;
&div{
border-color:red;
}
}
<span class="i-style"></span>
<div class="i-style"></div>
In that case div should have red color border.
If I understand your question maybe this help you:
SASS
.i-style{
border:solid 1px black;
#at-root div#{&}{
border-color:red;
}
}
OUTPUT
.i-style {
border: solid 1px black;
}
div.i-style {
border-color: red;
}
#Carlos, please try this. I hope this resolves your requirement
$selectorVar:'null' !default;
.i-style{
border:solid 1px black;
// build selector string
$selectorVar: div+& !global;
//$selectorVar: div+&, span+& !global; - for multiple selectors
}
#{$selectorVar} {
border-color:red;
}
This generates output like
.i-style {
border: solid 1px black;
}
div.i-style {
border-color: red;
}
Note: Generally, I don't pollute global namespace!. Accept this answer if it helps
Related
I'm writing a library and I'd like to style all buttons.
HTML
<div>
<p>Buttons</p>
<button>Button</button>
<button class="r1">Button</button>
</div>
<div>
<p>File inputs</p>
<input type="file" />
<input type="file" class="r1" />
</div>
SCSS
button,
input[type=file]::file-selector-button {
background: #81ecec;
border: 2px solid #00cec9;
&.r1{
background: red;
}
}
This code processes to:
button.r1,
input[type=file]::file-selector-button.r1 {
background: red;
}
[This is invalid and does not work]
Is there a mixin or method I can use so that I can place the classes on only the parent selector, without this getting out of hand? I intend to have multiple classes (primary, secondary, large, small) and I don't want to write:
button.r1,
input[type=file].r1::file-selector-button{
...
}
button.large,
input[type=file].large::file-selector-button{
...
}
button.small,
input[type=file].small::file-selector-button{
...
}
I can't figure out a good way of targeting the parent input[type="file"]
This codepen has the first example in it, and as it isn't valid CSS the background: red doesn't take effect:
https://codepen.io/EightArmsHQ/pen/VwxwPGM/139933ae274200149b84afdb726478c5?editors=1100
Attempt 1
At the moment I am using a mixin like so:
#mixin button{
background: var(--button-primary);
color: #fff;
text-decoration: none;
border: none;
display: inline-block;
padding: 4px 8px;
}
#mixin button-r1{
border-radius: 3px;
}
button,
.button,
input[type="submit"],
input[type="reset"]{
#include button;
&.r1{
#include button-r1;
}
}
input[type=file]{
&::file-selector-button{
#include button;
}
&.r1::file-selector-button{
#include button-r1;
}
}
The benefit is that I don't need to repeat the same styles over and over, however I feel like there must be a better way of creating a mixin that interpolates a class somehow.
Attempt 2
Using the classname as an argument works well, however I lose the ability to nest the rules, which is a shame and one of my favourite parts of SCSS.
#mixin buttonAndFileInputs($classname: "") {
button#{$classname},
.button#{$classname},
input[type="submit"]#{$classname},
input[type="reset"]#{$classname},
input[type="file"]#{$classname}::file-selector-button {
#content;
}
}
#include buttonAndFileInputs {
background: var(--button-primary);
color: #fff;
}
#include buttonAndFileInputs(".r1") {
border-radius: 3px;
}
I'm not 100% clear on what you're trying to do
But I think if you edit your codepen to this scss
button,
input[type=file] {
background: #81ecec;
border: 2px solid #00cec9;
&.r1{
background: red;
}
}
::file-selector-button {
background: inherit;
border: inherit;
}
that will get what you're looking for
Edit to add explanation:
This will make the file-selector-button follow the background and border properties of the input[type=file].
This means that the file-selector-button will match the rest of the input background.
I've created CSS for a LineEdit.
LineEdit.cpp
void MyLineEdit::initStyleSheet()
{
QString css = Css::instance().css( m_element );
setProperty( "style", "normal" );
setStyleSheet( css );
}
I have a separate .css file for style:
MyLineEdit.css
....
MyLineEdit[style="Normal"]:focus
{
border: 1px solid red;
}
MyLineEdit[style="Normal"]:disabled
{
border: 1px solid gray;
background: gray;
}
Now there is one weird requirement: MyLineEdit should have a method called setNoFrame, in this function we set one more property for it, and this property is valid for only state disabled.
This is what I did:
MyLineEdit::setNoFrame()
{
setProperty("noFrame","true");
initSyleSheet();
}
And this is my updated .css data
....
MyLineEdit[style="Normal"]:focus
{
border: 1px solid red;
}
MyLineEdit[style="Normal"]:disabled
{
border: 1px solid gray;
background: gray;
}
MyLineEdit[style="Normal", noFrame="true"]:disabled
{
border: none;
background: gray;
}
It doesn't work as I expected, the border is still there for state disabled and noFrame = true. Do I have mistake in combining properties for CSS above?
You're really, really close. Try
MyLineEdit[style="Normal"][noFrame="true"]:disabled
{
border: none;
background: gray;
}
From the CSS2 docs (which Qt StyleSheets supports):
Multiple attribute selectors can be used to refer to several attributes of an element, or even several times to the same attribute.
Here, the selector matches all SPAN elements whose "hello" attribute has exactly the value "Cleveland" and whose "goodbye" attribute has exactly the value "Columbus":
span[hello="Cleveland"][goodbye="Columbus"] { color: blue; }
Is there a keyword like currentcolor which allows us to get the color of a class in its default state?
For example, I'm trying to create a re-useable button style, and currentcolor keyword helps a lot until I try to create the :hovered state.
.outline-btn {
background-color: transparent;
border: 1px solid currentColor;
padding: 0.5em 1.5em;
}
.rounded-btn {
border-radius: 50px;
}
The default state looks the way we want and changing the color or the font-size would also adjust the rest of the properties.
But we want the :hovered state to invert the colors (white text and orange background in this case)
.outline-btn:hover, .outline-btn:active, .outline-btn:focus {
background-color: currentcolor;
color: white;
}
But since in this state the color becomes white, everything else also turns white.
Is there a way that we can achieve the behavior that we want without having to create multiple classes for the different button styles that we want?
Desired effect on hover:
Also I forgot to mention that I am using SCSS if that helps.
Thanks for your time :)
If you think about it, you're essentially wanting currentColor to act as a variable -- to hold a constant value. The upcoming CSS variables will help with this, but until they're better supported, we have Sass variables.
By defining the colors as variables you can write them out very verbosely and specifically, but only have to change the color in one place when needed.
$btn-color: red;
$btn-bg: transparent;
.outline-btn {
background-color: $btn-bg;
border: 1px solid $btn-color;
padding: 0.5em 1.5em;
color: $btn-outline-color;
&:hover,
&:active,
&:focus {
background-color: $btn-outline-color;
color: $btn-outline-bg;
}
}
You could go a step further and have those variables set to equal previously set variables you're using for the body/html color background, e.g., $bg-bg: $body-bg; $btn-color: $text-color;. I love currentColor as well and this isn't as clean as that, but it might be more appropriate in this case.
You can then build this out as a mixin as user6292372 noted. Something like:
#mixin buttonBuilder($color, $bg) {
background-color: $bg;
border: 1px solid $color;
color: $color;
&:hover {
background-color: $color;
color: $bg;
}
}
...
.outline-btn {
#include button-builder($btn-color, $btn-bg);
}
Then you can easily make multiple variants.
this can't be done with css only
if you use helpers like SCSS or Less you could make yourself a mixin where you only insert the color you want as a parameter.
but you would still have to make several classes (as many as you need different colors) but can reuse your mixin within like this (scss example):
#mixin invertHover($color) {
background-color: transparent;
border: 1px solid $color;
color: transparent;
&:hover {
background-color: $color;
border: 1px solid transparent;
color: $color;
}
}
.blue-box { #include invertHover('blue'); }
.black-box { #include invertHover('#000000'); }
For example, which is better:
Method 1 (name classes separately):
/* CSS */
.textbox-red,
.textbox-green {
padding: 10px;
border: 1px solid #CCC;
border-radius: 10px;
}
.textbox-red { color: #900; }
.textbox-green { color: #3c3; }
/*HTML*/
<div class="textbox-red"></div>
<div class="textbox-green"></div>
OR ------------
Method 2 (chain classes):
/* CSS */
.textbox {
padding: 10px;
border: 1px solid #CCC;
border-radius: 10px;
}
.textbox.text-red { color: #900; }
.textbox.text-green { color: #3c3; }
/*HTML*/
<div class="textbox text-red"></div>
<div class="textbox text-green"></div>
What is a better practice among the two?
My opinion is that you should use modular css -
You could also combine classes instead of linking them:
/*CSS*/
.textbox {
padding: 10px;
border: 1px solid #CCC;
border-radius: 10px;
}
.text-red { color: #900; }
.text-green { color: #3c3; }
/*HTML*/
<div class="textbox text-red"></div>
<div class="textbox text-green"></div>
That way you can reuse the red and green colors in cases when you want to have a red background without a textbox. This way you can re-use your code more and you have a loose coupling between your textbox and text-color
I personally would go with method 2. That way you can swap out text-red or text-green easily for text-blue or text-yellow and still keep the underlying style for your text. Basically, method 2 allows for more flexibility and maintainability, IMHO.
In my experience the modular approach gives you the most flexibility. The modular css pattern is also used in Twitter Bootstrap where flexibility is very important.
First better if you making style for IE6, because this browser dont support second method.
Is there a way to add scope to sass variables?
I want to be able to attach a class to my body element. The class will refer to a set of colours that the rest of the stylesheets can access.
I have tried:
#mixin theme_one{
$color: #000;
}
.theme_one{
#include theme_one;
}
and
.theme_one{
$color: #000;
}
I've just come across the same issue myself. I wanted to have different colour themes for different sections of my site.
Using a mixin seems like the best way to go. It's nicely DRY, and easy to use. The trick is not setting your colours in your main styles blocks, but rather using only the mixin for this.
I've set up the theme colours as variables at the top so they can be edited nicely, and I've set them as lists so that multiple values can be passed without hordes of variable being defined.
So:
// Variable Definitions
$defaultColor: black white grey;
$color2: blue green brown;
$color3: red white blue;
#mixin colorSet($color: $defaultColor) {
$link: nth($color, 1);
$border: nth($color, 2);
$background: nth($color, 3);
border-color: $border;
background-color: $background;
.column {
border-color: lighten($border, 10%);
}
a {
color: $link;
&:hover {
color: darken($link, 15%);
}
}
}
// Default colours
body {
#include colorSet();
}
// Scoped colours
.my-theme-3 {
#include colorSet($color3);
}
.my-theme-2 {
#include colorSet($color2);
}
Will produce something like this:
body {
border-color: white;
background-color: grey; }
body .column {
border-color: white; }
body a {
color: black; }
body a:hover {
color: black; }
.my-theme-3 {
border-color: white;
background-color: blue; }
.my-theme-3 .column {
border-color: white; }
.my-theme-3 a {
color: red; }
.my-theme-3 a:hover {
color: #b30000; }
.my-theme-2 {
border-color: green;
background-color: brown; }
.my-theme-2 .column {
border-color: #00b300; }
.my-theme-2 a {
color: blue; }
.my-theme-2 a:hover {
color: #0000b3; }
Edit: Updated to use default mixin values.
In your case no need to use mixin, If you have set of many styles then use mixin,
ie. if you have
#mixin theme_one{
$color: #000;
height: 50px;
}
then use Mixin
otherwise for single property use only variable
$color: #fff;
.some_class01{
color: $color;
background: $color;
}
.some_class22{
border-color: $color;
}
IMP: Variable should assign at the top of your code, it means don't use it after/below where you assigned it :)
Not sure if this is what you are looking for. It looks like you may have tried something similar to this,
which should probably work. (it may just be a matter of using !default)
Your body tag with a class on it..
<body class="theme_one">
</body>
Sass variables defined in stylesheet..
//THEME ONE VARIABLES
.theme_one{
$borderColor:#333 !default;
$fontColor:#999 !default;
}
//THEME TWO VARIABLES
.theme_two{
$borderColor:#CCC !default;
$fontColor:#000 !default;
}
Pre-existing CSS which will be overwritten depending on which class is used on the body tag.
h1.someheader {
color:$fontColor;
border-bottom:1px solid;
border-color:$borderColor;
}
Otherwise you could maybe try something like this. It looks like you may have tried something similar, however there seems to be an error with your mixin ... see note below.
//mixin used to set variables for properties
#mixin themeOne($fontColor,$borderColor) {
color:$fontColor;
border-color:$borderColor;
}
#include themeOne(#000,#CCC);
Pre-existing CSS
h1.someheader {
color:$fontColor
border-color:$borderColor;
border-bottom:1px solid;
}
Also note in your mixin example you are using $color:#000; ... This won't be interpreited properly as it should be color:#000; You can't use variables as selectors
unless you do something like #{$color}:#000;
I haven't quite tested this yet, so some things might need to be adjusted. If this doesn't solve your problem I hope it at least gives you some ideas.