jQuery-查找具有给定数量的同级元素

Question

I need to be able to search through each level of descendants of a JQuery object, and return any collections of siblings that may have a given number of members at that hierarchy level, starting with the objects most distant descendants, and working its way back up.

So given some HTML:

<div class="search-me">
  <div> <!-- DON'T FIND THIS -->
    <p>FIND ME</p>
    <p>FIND ME</p>
    <p>FIND ME</p>
  </div>
  <div> <!-- DON'T FIND THIS -->
    <span></span>
    <span></span>
  </div>
  <div> <!-- DON'T FIND THIS -->
    <div>FIND ME</div>
    <div>FIND ME</div>
    <div>FIND ME</div>
  </div>
</div>

I need the following pseudo code, to return the items labelled 'FIND ME', and nothing else...

$('.search-me').findGroupWithSize(3);

Also, note that there are three div's containing the <p> , <span> , and <div> tags, which should not be returned. So it should return a collection of the lowest level set of elements, whose siblings total a given amount.

Answer 1

The below example finds the groups of 3 from the HTML you posted.

 var siblingsGroupTarget = 3; var foundIndex = 0; var elementArray = []; $(".search-me").find("*").each(function(index){ // Conditions var hasChilds = $(this).children().length; var hasSiblings = $(this).siblings().length; var itsSiblings = $(this).siblings(); var itsSiblingsHasChilds = itsSiblings.children().length; console.log( $(this)[0].tagName+": "+$(this).siblings().length ); if( hasChilds == 0 && ( hasSiblings == siblingsGroupTarget-1 && itsSiblingsHasChilds == 0 ) ){ elementArray.push( $(this) ); } }); elementArray.reverse(); for(i=0;i<elementArray.length;i++){ foundIndex++; elementArray[i].addClass("FOUND").append(" I was found #"+foundIndex); }; 

 .FOUND{ border:3px solid red; width:12em; } 

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div class="search-me"> <div> <!-- DON'T FIND THIS --> <p>FIND ME</p> <p>FIND ME</p> <p>FIND ME</p> </div> <div> <!-- DON'T FIND THIS --> <span></span> <span></span> </div> <div> <!-- DON'T FIND THIS --> <div>FIND ME</div> <div>FIND ME</div> <div>FIND ME <ul> <!-- DON'T FIND THIS --> <li>FIND ME</li> <li>FIND ME</li> <li>FIND ME</li> </ul> </div> </div> </div> 

Answer 2

You can use selector ".search-me > *" passed to jQuery() to get direct descendants of .search-me element, .toArray() to convert jQuery object to array, Array.prototype.reverse() to reverse the collection of elements, .filter() to check if the child elements .length is N , wrap array methods in jQuery() with selector "> *" to return child elements of matched elements

 const N = 3; let res = $("> *", $(".search-me > *").toArray().reverse() .filter(function(el) {return $("> *", el).length === N})); console.log(res); res.css("color", "purple") 

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"> </script> <div class="search-me"> <div> <!-- DON'T FIND THIS --> <p>FIND ME</p> <p>FIND ME</p> <p>FIND ME</p> </div> <div> <!-- DON'T FIND THIS --> <span>DON'T FIND ME</span> <span>DON'T FIND ME</span> </div> <div> <!-- DON'T FIND THIS --> <div>FIND ME</div> <div>FIND ME</div> <div>FIND ME</div> </div> </div>