Get hidden input value from database after select form

Question

i have a select input that get options from database retailer_id, what i need it, when the admin select a retailer on this select input, i'll have a new one with the retailer_slug, and then, when the admin create or update, will sen to database the retailer_id and the retailer_slug on hidden input. my code is:

EDITED

        <select class="textbox2" id="retailer_id" name="retailer_id">
<option value="">--- Please select store ---</option>
<?php
    $sql_retailers = smart_mysql_query("SELECT * FROM cashbackengine_retailers WHERE status='active' ORDER BY title ASC");
    while ($row_retailers = mysql_fetch_array($sql_retailers))
    {
        if ($retailer_id == $row_retailers['retailer_id']) $selected = " selected=\"selected\""; else $selected = "";
        echo "<option data-slug=\"".$row_retailers['slug_title']."\" value=\"".$row_retailers['retailer_id']."\"".$selected.">".$row_retailers['title']."</option>";
    }
?>
</select>

<input type="text" name="slug" id="slug" value=""/>
         <script type="text/javascript">
        $(document).ready(function() {
        $('#retailer_id').on('change', function() {
var $selected = $('#retailer_id option:selected');
$('input[name=slug]').val($selected.data('slug'));
});
});
        </script>

So, retailer_slug is in the same table (cashbackengine_retailers)? — musicnothing
– musicnothing, Commented Nov 12, 2013 at 18:59
Yes, same table, just different collumn. So, when user select a retailer, i need to grab on the hidden input the retailer_slug, and then on click save, send the hidden input togheter with retailer_id — William Xavier
– William Xavier, Commented Nov 12, 2013 at 19:06

musicnothing · Accepted Answer · 2013-11-12 19:16:41Z

1

If I understand correctly, this is one thing you could do:

PHP

<select class="textbox2" id="retailer_id" name="retailer_id">
    <option value="">--- Please select store ---</option>
    <?php
        $sql_retailers = smart_mysql_query("SELECT * FROM cashbackengine_retailers WHERE status='active' ORDER BY title ASC");
        while ($row_retailers = mysql_fetch_array($sql_retailers))
        {
            if ($retailer_id == $row_retailers['retailer_id']) $selected = " selected=\"selected\""; else $selected = "";
            echo "<option data-slug=\"".$row_retailers['retail_slug']."\" value=\"".$row_retailers['retailer_id']."\"".$selected.">".$row_retailers['title']."</option>";
        }
    ?>
</select>
<input type="hidden" name="retailer_slug"/>

jQuery

$('#retailer_id').on('change', function() {
    var $selected = $('#retailer_id option:selected');
    $('input[name=retailer_slug]').val($selected.data('slug'));
});

So basically, the idea here is that you save the retailer_slug value as a data- attribute on each <option>. Then when the selected option is changed, the retailer_slug value is copied over to the hidden input.

answered Nov 12, 2013 at 19:16

musicnothing

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6 Comments

William Xavier Over a year ago

I'll test it right now, lets see if solves the problem, i'll get in touch, thanks :)

William Xavier Over a year ago

Let me ask, how can i test, like, output on the page to check if the hidden input its beeing populated with the slug?

musicnothing Over a year ago

If you have Chrome or Firefox you can right-click on use "Inspect Element." Or you could change the input type to text and then it'll be visible.

William Xavier Over a year ago

Just did and returned that error: Uncaught TypeError: Object #<Object> has no method 'on'

William Xavier Over a year ago

Ok, changed to $('#retailer_id').change(function() { var $selected = $('#retailer_id option:selected'); $('input[name=slug_title]').val($selected.data('slug')); }); But still dont return the slug value, what i'm doing wrong?

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