Selecting only first level element, not child elements with the same element name

Question

How would I select only the first level .block and not any of the children?

$(\'.block:not(\"Children of this here\")\') <--

Answer 1

Try this:

$("div.block:first").<what you want to do>

edit: removed the spaces. thanks :-)....now it should be good.

HTH

Answer 2

You can use the :first selector to select only the first matching .block element:

$('.block:first')

This works because jQuery matches elements in document order. The outermost .block element will be the first element matched by .block, and :first will filter to only return it.

Note that :first is not the same as :first-child.

EDIT: In response to your update, you can write the following, which will only work if all of the elements are nested three deep:

$('.block:note(:has(.block .block))')

You can write a more robust solution using a function call:

$('.block').not(function() { return $(this).closest('.block').length; })

This will find all .block elements, then remove any matched elements that have an ancestor matching .block. (You can replace closest with parent if you want to).

Answer 3

$('#parent').children('.block');

See .children() jQuery reference.

Answer 4

If that sample markup has a parent element, for example below. If not, the parent will be body if it is valid HTML.

HTML

<div id="parent">
<div class="block">
  <div class="block">
    <div class="block">

    </div>
  </div>
</div>
</div>

jQuery

$('#parent > .block').css({ border: '1px solid red' });