Can I combine :nth-child() or :nth-of-type() with an arbitrary selector?

Question

This is a very common problem that arises due to a misunderstanding of how :nth-child(An+B) and :nth-of-type() work.

In Selectors level 3, the :nth-child() pseudo-class counts elements among all of their siblings under the same parent. It does not count only the siblings that match the rest of the selector. Similarly, the :nth-of-type() pseudo-class counts siblings sharing the same element type, which refers to the tag name in HTML, and not the rest of the selector.

This also means that if all the children of the same parent are of the same element type, for example in the case of a table body whose only children are tr elements or a list element whose only children are li elements, then :nth-child() and :nth-of-type() will behave identically, i.e. for every value of An+B, :nth-child(An+B) and :nth-of-type(An+B) will match the same set of elements.

In fact, all simple selectors in a given compound selector, including pseudo-classes such as :nth-child() and :not(), work independently of one another, rather than looking at the subset of elements that are matched by the rest of the selector.

This also implies that there is no notion of order among simple selectors within each individual compound selector¹, which means for example the following two selectors are equivalent:

table.myClass tr.row:nth-child(odd)
table.myClass tr:nth-child(odd).row

Translated to English, they both mean:

Select any tr element that matches all of the following independent conditions:

• it is an odd-numbered child of its parent;
• it has the class "row"; and
• it is a descendant of a table element that has the class "myClass".

^{(you'll notice my use of an unordered list here, just to drive the point home)}

Selectors level 4 seeks to rectify this limitation by allowing :nth-child(An+B of S)² to accept an arbitrary selector argument S, again due to how selectors operate independently of one another in a compound selector as dictated by the existing selector syntax. So in your case, it would look like this:

table.myClass tr:nth-child(odd of .row)

Of course, being a brand new proposal in a brand new specification, this probably won't see implementation until a few years down the road.

In the meantime, you'll have to use a script to filter elements and apply styles or extra class names accordingly. For example, the following is a common workaround using jQuery (assuming there is only one row group populated with tr elements within the table):

$('table.myClass').each(function() {
  // Note that, confusingly, jQuery's filter pseudos are 0-indexed
  // while CSS :nth-child() is 1-indexed
  $('tr.row:even').addClass('odd');
});

With the corresponding CSS:

table.myClass tr.row.odd {
  ...
}

If you're using automated testing tools such as Selenium or scraping HTML with tools like BeautifulSoup, many of these tools allow XPath as an alternative:

//table[contains(concat(' ', @class, ' '), ' myClass ')]//tr[contains(concat(' ', @class, ' '), ' row ')][position() mod 2)=1]

Other solutions using different technologies are left as an exercise to the reader; this is just a brief, contrived example for illustration.

---

¹_{If you specify a type or universal selector, it must come first. This does not change how selectors fundamentally work, however; it's nothing more than a syntactic quirk.}

²_{This was originally proposed as :nth-match(), however because it still counts an element relative only to its siblings, and not to every other element that matches the given selector, it has since as of 2014 been repurposed as an extension to the existing :nth-child() instead.}

Not really..

quote from the docs

The :nth-child pseudo-class matches an element that has an+b-1 siblings before it in the document tree, for a given positive or zero value for n, and has a parent element.

It is a selector of its own and does not combine with classes. In your rule it just has to satisfy both selector at the same time, so it will show the :nth-child(even) table rows if they also happen to have the .row class.

nth-of-type works according to the index of same type of the element but nth-child works only according to index no matter what type of siblings elements are.

For example

<div class="one">...</div>
<div class="two">...</div>
<div class="three">...</div>
<div class="four">...</div>
<div class="five">...</div>
<div class="rest">...</div>
<div class="rest">...</div>
<div class="rest">...</div>
<div class="rest">...</div>
<div class="rest">...</div>

Suppose in above html we want to hide all the elements having rest class.

In this case nth-child and nth-of-type will work exactly same as all the element are of same type that is <div> so css should be

.rest:nth-child(6), .rest:nth-child(7), .rest:nth-child(8), .rest:nth-child(9), .rest:nth-child(10){
    display:none;
}

OR

.rest:nth-of-type(6), .rest:nth-of-type(7), .rest:nth-of-type(8), .rest:nth-of-type(9), .rest:nth-of-type(10){
    display:none;
}

Now you must be wondering what is the difference between nth-child and nth-of-type so this is the difference

Suppose the html is

<div class="one">...</div>
<div class="two">...</div>
<div class="three">...</div>
<div class="four">...</div>
<div class="five">...</div>
<p class="rest">...</p>
<p class="rest">...</p>
<p class="rest">...</p>
<p class="rest">...</p>
<p class="rest">...</p>

In the above html the type of .rest element is different from others .rest are paragraphs and others are div so in this case if you use nth-child you have to write like this

.rest:nth-child(6), .rest:nth-child(7), .rest:nth-child(8), .rest:nth-child(9), .rest:nth-child(10){
    display:none;
}

but if you use nth-of-type css can be this

.rest:nth-of-type(1), .rest:nth-of-type(2), .rest:nth-of-type(3), .rest:nth-of-type(4), .rest:nth-of-type(5){
    display:none;
}

As type of .rest element is <p> so here nth-of-type is detecting the type of .rest and then he applied css on the 1st, 2nd, 3rd, 4th, 5th element of <p>.

You may be able to do that with xpath. something like //tr[contains(@class, 'row') and position() mod 2 = 0] might work. There are other SO questions expanding on the details how to match classes more precisely.

Here is your answer

<!DOCTYPE html>
<html lang="en">
<head>
  <meta charset="UTF-8">
  <meta name="viewport" content="width=device-width, initial-scale=1">
  <title>TEST</title>

  <style>

    .block {
      background: #fc0;
      margin-bottom: 10px;
      padding: 10px;
    }
    /* .large > .large-item:nth-of-type(n+5) {
      background: #f00;
    } */

    .large-item ~ .large-item ~ .large-item ~ .large-item ~ .large-item {
      background: #f00;
    }

  </style>
</head>
<body>

<h1>Should be the 6th Hello Block that start red</h1>
<div class="small large">
  <div class="block small-item">Hello block 1</div>
  <div class="block small-item large-item">Hello block 2</div>
  <div class="block small-item large-item">Hello block 3</div>
  <div class="block small-item large-item">Hello block 4</div>
  <div class="block small-item large-item">Hello block 5</div>
  <div class="block small-item large-item">Hello block 6</div>
  <div class="block small-item large-item">Hello block 7</div>
  <div class="block small-item large-item">Hello block 8</div>
</div>

</body>
</html>