How to work with the first 4 elements that have a specific attribute? It is necessary to work with elements, with style = 'display: none' html structure:

<div class="wrapper"> <div>item</div> <div>item</div> <div style="display: none;">item</div> <div style="display: none;">item</div> <div style="display: none;">item</div> <div style="display: none;">item</div> <div style="display: none;">item</div> <div style="display: none;">item</div> <div style="display: none;">item</div> <div style="display: none;">item</div> </div>

    2 answers 2

     //получаем все элементы со значением атрибута style 'display: none;' var elements = jQuery('[style = "display: none;"]'); //выбираем первые четыре элемента for (var i = 0; i < 4; i++) { console.log(elements[i]); } 
     <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script> <div class="wrapper"> <div>item</div> <div>item</div> <div style="display: none;">item 1</div> <div style="display: none;">item 2</div> <div style="display: none;">item 3</div> <div style="display: none;">item 4</div> <div style="display: none;">item 5</div> <div style="display: none;">item 6</div> <div style="display: none;">item 7</div> <div style="display: none;">item 8</div> </div> 

       var elements = $(".wrapper div").filter(function() { return $(this).css('display') == 'none'; }).slice(0,4); elements.css("display","block").html("hello") 
       <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div class="wrapper"> <div>item</div> <div>item</div> <div style="display: none;">item</div> <div style="display: none;">item</div> <div style="display: none;">item</div> <div style="display: none;">item</div> <div style="display: none;">item</div> <div style="display: none;">item</div> <div style="display: none;">item</div> <div style="display: none;">item</div> </div>