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Home/ Questions/Q 8657805
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T15:36:44+00:00 2026-06-12T15:36:44+00:00

I had three drop down menus. Depending on items selected on two lists, the

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I had three drop down menus. Depending on items selected on two lists, the third list should be filled out (ajax post).

This is the html code:

Source Language:
<select name='source_lang' id='source_lang' $dis size='6'>
<option value='en'>EN</option>
<option value='sq'>SQ</option>
.....
</select>

Target Language:
<select name='targ_lang' id='targ_lang' $dis size='6'>
<option value='en'>EN</option>
<option value='sq'>SQ</option>
.....
</select>


Supplier:
<select name='supplier_id' id='supplier_id'>
<option value='1'>Supplier 1</option>
<option value='2'>Supplier 2</option>
</select>

On target and source language change, the supplier select list should be filled out.

Anybody can help me with the jquery? I need to ajax post the source, target language values and as response fill out the supplier select list with the data.

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  1. Editorial Team
    Editorial Team
    2026-06-12T15:36:46+00:00Added an answer on June 12, 2026 at 3:36 pm

    Is this something you’re looking for?

    $('#targ_lang, #source_lang').change(function(){
           $.ajax({
              url : '',
              method : 'post',
              type : 'json',
              data : {
                select1 : $('#targ_lang').val(),
                select2 : $('#source_lang').val()
              },
              complete : function(result){
                var options = $.parseJSON(result);
                $('#supplier_id').html("");
                for(i=0; i < options.length; i++) {
                    $('#supplier_id').append(
                         '<option value="'+ options[i] + '">' + 'Supplier' + options[i] + '</option>'
                    );
                }
              });
          });
    });
    

    On the PHP side you need to send your result like this:

    $array = new Array(1,2,3,4); // supplier IDs
    echo json_encode($array);
    
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