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Home/ Questions/Q 778981
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Editorial Team
  • 0
Editorial Team
Asked: May 14, 20262026-05-14T19:51:15+00:00 2026-05-14T19:51:15+00:00

Hi I want to count LI which does not have the UL , for

  • 0

Hi I want to count LI which does not have the UL , for the first level only,
but when I count this it shows size 4 instead of 2, its count the inner LI also.

<div class="navigation-container">
    <ul class="first-level">
      <li><a href="#">Link 1</a></li>
      <li><a href="#">Link 2</a>
        <ul>
          <li><a href="#">Link2.1</a></li>
          <li><a href="#">Link2.2</a>
            <ul>
                <li><a href="#">Link 2.2.1</a></li>
            </ul>
          </li>
        </ul>
      </li>
      <li><a href="#">Link </a></li>
    </ul>  
  </div>

jQuery for this.

jQuery(document).ready(function(){

  var nosubnav = jQuery('.first-level li:not(:has(ul))');
  var nosubnavsize = jQuery('.first-level li:not(:has(ul))').size();
  jQuery(nosubnav).css('border' , '1px solid red');
  alert('List item which does not have submenu  '+nosubnavsize);

});

Link for the testing link text on JSBin,

thanks

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T19:51:15+00:00Added an answer on May 14, 2026 at 7:51 pm

    You can use the child selector > to target only child elements directly under the parent.

    jQuery(document).ready(function(){
    
      var nosubnav = jQuery('.first-level > li:not(:has(ul))');
      var nosubnavsize = jQuery('.first-level > li:not(:has(ul))').size();
      jQuery(nosubnav).css('border' , '1px solid red');
      alert('List item which does not have submenu  '+nosubnavsize);
    
    });
    

    This will return the count of 2. You can also optimise this slightly by reusing your stored selection of the target li (stored in nosubnav):

    jQuery(document).ready(function(){
    
      var nosubnav = jQuery('.first-level > li:not(:has(ul))');
      nosubnav.css('border' , '1px solid red');
      alert('List item which does not have submenu  '+nosubnav.length);
    
    });
    

    This will cut out the overhead of querying the DOM a second time.

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