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Home/ Questions/Q 7993403
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T13:50:31+00:00 2026-06-04T13:50:31+00:00

I’m stuck on something that I think should be easy… I’ve got a select

  • 0

I’m stuck on something that I think should be easy… I’ve got a select that on change gets the id value and im tring to hide all and show th ones that with the same id value
so it works just with the first
any help me please

<select name="" id="estado_list">
 <option value="0">-</option>
 <option value="1">Nuevo León</option>
 <option value="2">Puebla</option>
</select>


<ul style="list-style: none outside none;">
 <li class="forli" id="1">data</li>
 <li class="forli" id="1">data</li>
 <li class="forli" id="1">data</li>
 <li class="forli" id="1">data</li>
 <li class="forli" id="1">data</li>
 <li class="forli" id="1">data</li>
 <li class="forli" id="2">data</li>
</ul>






 $(document).ready(function(){  
        $("#estado_list").change(function(){  
         var id = $(this).val();

          if (id == 0 ){$('.forli').show();}
       else{$('.forli').hide();
            $('li').each(function(){
             $('#'+ id).show();

         });
         }
        }); 
    });  
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-04T13:50:32+00:00Added an answer on June 4, 2026 at 1:50 pm

    Id fields should be unique and will cause problems when accessed from the DOM object. Try using the class field instead.

    <select name="" id="estado_list">
     <option value="list0">-</option>
     <option value="list1">Nuevo León</option>
     <option value="list2">Puebla</option>
    </select>
    
    
    <ul style="list-style: none outside none;">
     <li class="forli list1">data</li>
     <li class="forli list1">data</li>
     <li class="forli list1">data</li>
     <li class="forli list1">data</li>
     <li class="forli list1">data</li>
     <li class="forli list1">data</li>
     <li class="forli list2">data</li>
    </ul>
    
    
     $(document).ready(function(){  
       $("#estado_list").change(function(){  
          $('.forli').hide().filter("."+$(this).val()).show();
       });
     });
    
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