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Home/ Questions/Q 450029
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Editorial Team
  • 0
Editorial Team
Asked: May 12, 20262026-05-12T21:49:42+00:00 2026-05-12T21:49:42+00:00

I have a select field. I must fill with options taken from a mysql

  • 0

I have a select field. I must fill with options taken from a mysql table.
Here is some little php code I have done using codeigniter framework

$idcateg = trim($this->input->post('idcategory'));
$array1 = array(
    'result' => $idcateg
);
echo json_encode($array1);

Now, the jQuery call…

$.post("<?=base_url()?>index.php/rubro/list_ajax/", { 
    'idcategory' : idc },
    function(data){
        alert(data.result);
    }, "json");

The code works fine. When I call the post, I get the categoryid as a result.
Now, I should modify the code above, so I can do:

  • post the ajax call sending the category id. this is done
  • get subcategories for this category, and build the array *
  • json_encode the array and echo *
  • get results back in jQuery ajax call, decode and build the < select > field *

The array should be built with each element having a sub-array with id and name, right?
Thanks a lot in advance for any help

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-12T21:49:42+00:00Added an answer on May 12, 2026 at 9:49 pm

    It’s not much different.

    $idcateg = trim($this->input->post('idcategory'));
    $result = array();
    $id = mysql_real_escape_string($idcateg);
    $res = mysql_query("SELECT * FROM subcategories WHERE category = $id");
    while ($row = mysql_fetch_array($res)) {
      $result[] = array(
        'id' => $row['subcatid'],
        'desc' => $row['description'],
      );
    }
    echo json_encode($result);
    

    with:

    $.post("<?=base_url()?>index.php/rubro/list_ajax/", { 
      'idcategory' : idc },
      function(data) {
        var sel = $("#select");
        sel.empty();
        for (var i=0; i<data.length; i++) {
          sel.append('<option value="' + data[i].id + '">' + data[i].desc + '</option>');
        }
      }, "json");
    
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