CMellor Posted March 11, 2007 Share Posted March 11, 2007 Hello, I have been trying to develop a bit of code to allow me to update a database field when an onChange occures on a drop down box. For some odd reason, it doesn't work at all, also I might add I am using Prototype. Here's my code: <script language="Javascript" type="text/javascript"> // Update roster brand function changeBrand(value, id) { new Ajax.Updater('div', 'js/brand.php?show=' + value + '&id=' + id); } </script> <?php // MySQL connect include('../../include/functions.php'); // connect connect(); if(!isset($_GET['show'])) { mysql_query("UPDATE ibf_members SET mgroup = '".$_GET['show']."' WHERE id = '".$_GET['id']."'"); } // close MySQL mysql_close(); ?> echo(' <form action="" method="post" name="form"> <select name="brand" onchange="changeBrand(document.form.brand.options[document.form.brand.selectedIndex].value, '.$row['id'].'); return false"> '); // query $group_query = mysql_query("SELECT * FROM ibf_groups, ibf_members WHERE ibf_groups.g_id = ibf_members.id AND g_id IN(3, 4, 6, 7)"); // fetch data while($groups = mysql_fetch_array($group_query)) { $selected = $row['mgroup'] == $groups['g_id'] ? 'selected' : ''; echo(' <option value="'.$groups['g_id'].'" '.$selected.'>'.$groups['g_title'].'</option> '); } echo(' </select> </form> <div id="div"></div> '); Does anything look wrong? Link to comment https://forums.phpfreaks.com/topic/42263-ajaxphp-and-drop-down-menus/ Share on other sites More sharing options...
ober Posted March 12, 2007 Share Posted March 12, 2007 2 things: Try changing it to onChange (capital C) Or try using a different event. You're also not checking the result of the SQL for a valid result (nor are you cleaning the input). That's never good. Are you getting any JS errors otherwise? Link to comment https://forums.phpfreaks.com/topic/42263-ajaxphp-and-drop-down-menus/#findComment-205707 Share on other sites More sharing options...
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