Call PHP via AJAX

Question

My user needs to pick an option from a first <select> and according to his selection, a second <select> should offer different options.

You may have seen this when picking a State after picking a Country in many many websites.

I know there are hundred other questions like this but I have tried lots of their answers without any success.

So, here is where I am standing right now:

            <div class="col-md-2">
              <span class="input-group-addon" id="basic-addon1" style="border:solid 1px #CCC; border-radius:4px;">Juzgado: </span>
              <select class="form-control" id="juz" name="juz">
                <option value="1">Juzgado 1</option>
                <option value="2">Juzgado 2</option>
                <option value="3">Juzgado 3</option>
              </select>
            </div>

            <div class="col-md-2" id="nomDiv">
                  <span class="input-group-addon" id="basic-addon1" style="border:solid 1px #CCC; border-radius:4px;">Nominación: </span>
                  <select class="form-control" id="nom" name="id_juz_nom">

                  </select>
            </div>

Then... my AJAX

<script type='text/javascript'>
        $(document).ready(function(){
            $('select[name=juz]').on('change', function() {

                var juz = $(this).val() 

                $.ajax({
                url: "tomarNominaciones.php",
                type: "POST",
                data: { juz.value },

                success: function (response) {

                        document.getElementById('nom').innerHTML= response;

                }
            });
          });
        });

</script>

And finally this is my PHP:

    <?php 
$juz = $_POST['juz'];
include 'conectar.php';
mysql_select_db('juz_nom');
$query  = "SELECT * FROM juz_nom WHERE juzgado=\"".$juz."\"";
$result = mysql_query($query);
?>

<span class="input-group-addon" id="basic-addon1" style="border:solid 1px #CCC; border-radius:4px;">Nominación: </span>
<select class="form-control" id="nom" name="id_juz_nom">
<option>Seleccione Nominación</option>
<?php while ($row=mysql_fetch_array($result)) { ?>
<option value=<?php echo $row['id_juz_nom']?>><?php echo $row['nominacion']?></option>
<?php } ?>
</select>

I've tried many methods such as this one and this other one. Without any success.

So, to be clear, what I need to do is: populate my second select with as many options as my MySQL query returns.

juz.value will be undefined because juz is already a string. .val() already extracted the value. — castletheperson
– castletheperson, Commented Apr 17, 2016 at 21:11
Unfortunately, not right: A malicious user could just send a post request themselves with whatever value they like. — mmgross
– mmgross, Commented Apr 17, 2016 at 21:24
@GonHL Nope, a request could be made direct to your PHP page, or a user could modify the DOM so it sends something else.. Never trust client side input. Also use the @ to tag users. — chris85
– chris85, Commented Apr 17, 2016 at 21:26
Stop using the deprecated and as of PHP7 removed mysql_* functions. Migrate to PDO and start using Prepared Statements. — Charlotte Dunois
– Charlotte Dunois, Commented Apr 17, 2016 at 21:31

rojobo · Accepted Answer · 2016-04-18 00:51:56Z

0

Change

data: { juz.value },

to

data: { juz },

and echo the HTML

as:

    echo "<span class='input-group-addon' id='basic-addon1' style='border:solid 1px #CCC; border-radius:4px;'>Nominación: </span>
    <select class='form-control' id='nom' name='id_juz_nom'>
    <option>Seleccione Nominación</option>";
 while ($row=mysql_fetch_array($result)) {
    echo "<option value=$row['id_juz_nom']>$row['nominacion']</option>
    </select>";}

edited Apr 18, 2016 at 0:51

answered Apr 18, 2016 at 0:43

rojobo

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