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rated 0 times [  174] [ 7]  / answers: 1 / hits: 35546  / 9 Years ago, wed, september 2, 2015, 12:00:00

I have a few random blocks. Whenever a block falls in the new row, I am making it look different. When the user clicks on a button, I hide few blocks by display:none, and the problem occurs. The nth-child selector also counts hidden elements.



Is there way to ignore those specific blocks, so that again every row has a different style? This is an example of a similar scenario.





$('.hide-others').click(function () {
$('.css--all-photo').toggleClass('hidden');
})

.board-item--inner {
height:200px;
background:tomato;
text-align:center;
color:#fff;
font-size:33px;
margin-bottom:15px;
border:2px solid tomato;
}
@media (min-width:768px) and (max-width:991px) {
.board-item:nth-child(2n+1) .board-item--inner {
border:2px solid #000;
background:yellow;
color:#000;
}
}
@media (min-width:992px) and (max-width:1199px) {
.board-item:nth-child(3n+1) .board-item--inner {
border:2px solid #000;
background:yellow;
color:#000;
}
}
@media (min-width:1200px) {
.board-item:nth-child(4n+1) .board-item--inner {
border:2px solid #000;
background:yellow;
color:#000;
}
}

<link href=https://maxcdn.bootstrapcdn.com/bootstrap/3.3.5/css/bootstrap.min.css rel=stylesheet/>
<script src=https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js></script>
<div class=container>
<div class=form-group>
<button class=btn btn-info hide-others type=button>Hide others</button>
</div>
<div class=row>
<div class=col-xs-12 col-sm-6 col-md-4 col-lg-3 board-item photos-board-item>
<div class=board-item--inner>1</div>
</div>
<div class=col-xs-12 col-sm-6 col-md-4 col-lg-3 board-item photos-board-item>
<div class=board-item--inner>2</div>
</div>
<div class=col-xs-12 col-sm-6 col-md-4 col-lg-3 board-item photos-board-item css--all-photo>
<div class=board-item--inner>3</div>
</div>
<div class=col-xs-12 col-sm-6 col-md-4 col-lg-3 board-item photos-board-item>
<div class=board-item--inner>4</div>
</div>
<div class=col-xs-12 col-sm-6 col-md-4 col-lg-3 board-item photos-board-item>
<div class=board-item--inner>5</div>
</div>
<div class=col-xs-12 col-sm-6 col-md-4 col-lg-3 board-item photos-board-item css--all-photo>
<div class=board-item--inner>6</div>
</div>
<div class=col-xs-12 col-sm-6 col-md-4 col-lg-3 board-item photos-board-item>
<div class=board-item--inner>7</div>
</div>
<div class=col-xs-12 col-sm-6 col-md-4 col-lg-3 board-item photos-board-item css--all-photo>
<div class=board-item--inner>8</div>
</div>
<div class=col-xs-12 col-sm-6 col-md-4 col-lg-3 board-item photos-board-item>
<div class=board-item--inner>9</div>
</div>
<div class=col-xs-12 col-sm-6 col-md-4 col-lg-3 board-item photos-board-item>
<div class=board-item--inner>0</div>
</div>
<div class=col-xs-12 col-sm-6 col-md-4 col-lg-3 board-item photos-board-item>
<div class=board-item--inner>10</div>
</div>
</div>
<div>





Simply go through the snippet or EXTERNAL FIDDLE, and you'll get my question.



I am specifically looking for a pure CSS solution.
Please provide a fiddle for your answers! And I cannot remove those blocks permanently, my user has the option to filter files on button click, that is why hidden and shown scenario.


More From » jquery

 Answers
40

When the user clicks on a button, I hide few blocks by display:none,
and the problem occurs. The nth-child selector also counts hidden
elements.



Is there way to ignore those specific blocks, so that again every row
has different style?




The problem is that the nth-child() selector looks at all siblings under the same parent regardless of styling. It doesn't matter that you've applied display: none because CSS doesn't remove the element from the DOM, and therefore it's still a sibling.



From the spec:




6.6.5.2. :nth-child()
pseudo-class



The :nth-child(an+b) pseudo-class notation represents an
element
that has an+b-1 siblings before it in the document
tree
, for any positive integer or zero value of n, and has a parent
element. (emphasis mine)




In order for the nth-child rules you've declared to work after a user clicks to hide divs, you need to remove the hidden divs from the DOM, so they no longer exist as siblings.



In your question you request a CSS-only solution. But in your comments you say that the HTML is open for changes. You also use a bit of jQuery to hide elements.



With one small line of code added to your jQuery the problem can be solved:



$('.hidden').remove();


The .remove() method takes elements (and its descendants) out of the DOM. In this case it removes all elements with a class hidden.






CORRECTION



The problem with remove() is that elements taken from the DOM with this method can't be restored, and this breaks the toggle function.



Fortunately, jQuery offers an alternative: detach().




The .detach() method is the same as .remove(), except that
.detach() keeps all jQuery data associated with the removed
elements. This method is useful when removed elements are to be
reinserted into the DOM at a later time.




So if we replace the original code...



$('.hide-others').click(function () {
$('.css--all-photo').toggleClass('hidden');
})


...with this code...



var divs;

$('.photos-board-item').each(function(i){
$(this).data('initial-index', i);
});

$('.hide-others').on('click', function () {
if(divs) {
$(divs).appendTo('.row').each(function(){
var oldIndex = $(this).data('initial-index');
$('.photos-board-item').eq(oldIndex).before(this);
});
divs = null;
} else {
divs = $('.css--all-photo').detach();
}
});


... the grid works as intended. (code credit: @JosephMarikle)



DEMO



Now, regardless of which divs or how many are hidden, they can be toggled on and off without disrupting the visual design because the nth-child selector is counting only visible siblings. No changes to the CSS. No changes to the HTML.


[#65211] Monday, August 31, 2015, 9 Years  [reply] [flag answer]
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