I have a slider that's marked up like so:
<div class="slider wide">
//slider html in here
</div>
And another marked up like so:
<div class="slider narrow">
//slider html in here
</div>
Is it possible to reference each of these like this in my CSS file by in a way concatenating the class names:
.slider.wide { //css specific to the wide slider goes here }
.slider.narrow { //css specific to the wide slider goes here }
No, you make three classes .slider, where you put common slider css, and .narrow where you put narrow slider specific css, and .wide where you put wide slider specific css.
.slider { //css common among all sliders goes here }
.wide { //css specific to the wide slider goes here }
.narrow { //css specific to the narrow slider goes here }
Yes, .slider.narrow is valid. It's not exactly concatenating the class names, it's making two different class selectors and applying them to the same element. So .narrow.slider is also valid and will match the same elements.
The problem with using multiple class selectors against a single element is that is doesn't work in IE6. This browser will ignore all but the last class selector. So to support that browser you typically end up using something like class="slider wide-slider".
Related
I have multiple css classes that make up a button using SCSS.
.ghost-button {
// CSS goes here
}
.ghost-button-label {
// CSS goes here
}
.plus-circle {
//CSS goes here
}
Using Angular I can control the disabled state using the following feature.
[class.disabled]="booleanFlag"
I wanted to have a disabled state for this button with out having multiple disabled classes like so,
.ghost-button.disabled {
// CSS goes here
}
.ghost-button-label.disabled {
// CSS goes here
}
.plus-circle.disabled {
//CSS goes here
}
This is an example of what I am trying to do. This did not work for me.
.ghost-button .ghost-button-label .plus-circle-position .disabled {
//CSS goes here
}
Here is the markup I use for the button,
<div style="padding-top: 10px" (click)="handleClickAdd($event)">
<div class="ghost-button ghost-button-label icon-custom icon-margin plus-circle plus-circle-position" [class.disabled]="blockAdditions">
<div>
<div>Add</div>
</div>
</div>
</div>
Is there a way to do this? Thanks.
This doesn't work because it means each class is a descendant of the previous:
.ghost-button .ghost-button-label .plus-circle-position .disabled {
//CSS goes here
}
If you're trying to just select that one div with all four classes, just remove the spaces:
.ghost-button.ghost-button-label.plus-circle-position.disabled {
//CSS goes here
}
If you're trying to select any elements that have the disabled class plus one of the three other classes, then you use commas to separate the different combinations:
.ghost-button.disabled,
.ghost-button-label.disabled,
.plus-circle-position.disabled {
// CSS
}
Of course you could just select .disabled if you want this CSS applied to every element with the disabled class:
.disabled {
// CSS
}
Just be sure to take into account View Encapsulation. You may need to put this CSS in the global style file styles.css if this class exists in more than one component.
Just a note, you are not setting the disabled state here, you are adding a class with the name "disabled". disabled is a boolean attribute you can set via HTML, which you can then select with the pseudo-class :disabled.
button:disabled {
color: red
}
<button>Not Disabled</button>
<button disabled>Disabled</button>
If this is what you were actually trying to do then in Angular it would be:
[disabled]="booleanFlag"
You can target a disabled element with the :disabled pseudo-class https://developer.mozilla.org/en-US/docs/Web/CSS/:disabled
So depending on the relationship between your button/label/plus-circle you should be able to target those as well based on whether the button is disabled. For example, if the button and label are siblings you could do this:
.ghost-button:disabled,
.ghost-button:disabled + .ghost-button-label,
.ghost-button:disabled + .plus-circle {
// CSS goes here
}
That would only work if the label and circle were siblings that come after the button, if they are before the button, you can't select them that way.
how do you apply a specific css to second-page
HTML
<div className = "section-header second-page">SOME MESSAGE</div>
Assuming the above:
CSS
.section-header {
background-color: black,
}
i want to apply a different background color specifically to second-page that does not modify section-header.
If you want the styling to apply to any element with .second-page class you should use:
.second-page {
backgound-color: red,
}
If you want the styling to apply only to .section-header elements that also have .second-page class, then you should use:
.section-header.second-page {
backgound-color: red,
}
When there's no space between two classes, it means it refers to an element with both classes.
For more information on CSS selectors, please check
https://www.w3schools.com/cssref/css_selectors.asp
Your HTML should be:
<div class="section-header second-page">SOME MESSAGE</div>
Your CSS could be:
.second-page {
background-color: black,
}
You can mix multiple classes within the HTML or target them separately.
If you need to validate your HTML code you can use this free service:
https://validator.w3.org/#validate_by_input
I have a common element which contains articles, and want to treat all but the first child differently as follows:
.listing{
article{
// Some styles
}
article:not(:first-child){
// Some more styles
}
}
All well and good. However on some listings they should all be treated the same, so I don't want to include the article:not(:first-child) selector, it needs to be like the following:
.listing.alt{
article{
// Some styles
// Some more styles
}
}
How can I combine these two rules without repeating everything?
Ok I think I've figured it out using Sass:
.listing{
article{
// Generic Styles
}
&.alt article,
&:not(.alt) article:not(:first-child){
// More Styles
}
}
I also see that my original code example was a bit weird so I've updated it so it's a bit more correct.
HTML
<div class="listing">
<article>1</article>
<article>2</article>
<article>3</article>
</div>
<div class="listing alt">
<article>1</article>
<article>2</article>
<article>3</article>
</div>
CSS
.listing:not(.alt) article:not(:first-child) {color:gainsboro;}
Updated demo
I'm working on some CSS from a tutorial, a div has this class:
<div class="related products">
How can I reference it in the stylesheet?
The div actually has two classes, related and products. You can reference it in your stylesheet with either .related or .products, and it will pick up the styles from both of those rules. For example, with the following CSS, the text in the div in your question would appear red with font size 12:
.related { color:#ff0000 }
.products { font-size:12px }
If you want to select elements with both classes, use .related.products in your stylesheet. For example, add the following to the above example:
.related.products { font-weight:bold }
And the text in your div will receive all three rules, because it matches all 3 selectors. Here's a working example.
div.related.products is the general method
You reference it by div.related.products which literaly translates to "a div with class of related and class of products".
Or, you could reference it by using either class names, since it will catch both.
jsFiddle Example.
In the css, just put the name class of the div by doing this:
.related products {
/*styling to go here*/
}
Now any styling within the related products class will be applied to that div.
Is there a fast way in CSS to remove all of the styles applied to an element? For example, say a tab menu of some sort:
<div class='outer'>
<div id='d1'></div>
<div id='d2'></div>
<div id='d3'></div>
<div id='d4'></div>
</div>
The CSS is applied...
.outer { foo:blee; bar:blah; bas-bloo:snork; /*...long long list...*/ }
Now, I want #d3 (for example) to return to default styling, but I don't want to explicitly unset all of the parent styles:
#d3 { remove-styles:all } /* <- [I made this up, obviously] */
Pipe dream or possibility?
In CSS3, yes. You could use the negation pseudo-class:
.outer:not(#d3) { foo:blee; etc etc }
Too bad CSS3 support is a little lacking at the moment with most browsers...
With CSS level less than 3, you're screwed. Sorry.
No. Not feasibly possible. Just override it.