Saludos, como venia desarrollando un catalogo de productos me puse a usar chained select y al abrirlo funciona perfecto, pero me imprime este error que no se como solucionar
Notice: Undefined index: func in C:\Program Files\EasyPHP-5.3.3.1\www\surgimed\func.php on line 21
Notice: Undefined index: func in C:\Program Files\EasyPHP-5.3.3.1\www\surgimed\func.php on line 61
aqui les dejo el codigo del func.php
Código PHP:
<?php
//**************************************
// Page load dropdown results //
//**************************************
function getTierOne()
{
$result = mysql_query("SELECT DISTINCT tier_one FROM three_drops")
or die(mysql_error());
while($tier = mysql_fetch_array( $result ))
{
echo '<option value="'.$tier['tier_one'].'">'.$tier['tier_one'].'</option>';
}
}
//**************************************
// First selection results //
//**************************************
if($_GET['func'] == "drop_1" && isset($_GET['func'])) {
drop_1($_GET['drop_var']);
}
function drop_1($drop_var)
{
include_once('db.php');
$result = mysql_query("SELECT DISTINCT tier_two FROM three_drops WHERE tier_one='$drop_var'")
or die(mysql_error());
echo '<select name="drop_2" id="drop_2">
<option value=" " disabled="disabled" selected="selected">Choose one</option>';
while($drop_2 = mysql_fetch_array( $result ))
{
echo '<option value="'.$drop_2['tier_two'].'">'.$drop_2['tier_two'].'</option>';
}
echo '</select>';
echo "<script type=\"text/javascript\">
$('#wait_2').hide();
$('#drop_2').change(function(){
$('#wait_2').show();
$('#result_2').hide();
$.get(\"func.php\", {
func: \"drop_2\",
drop_var: $('#drop_2').val()
}, function(response){
$('#result_2').fadeOut();
setTimeout(\"finishAjax_tier_three('result_2', '\"+escape(response)+\"')\", 400);
});
return false;
});
</script>";
}
//**************************************
// Second selection results //
//**************************************
if($_GET['func'] == "drop_2" && isset($_GET['func'])) {
drop_2($_GET['drop_var']);
}
function drop_2($drop_var)
{
include_once('db.php');
$result = mysql_query("SELECT * FROM three_drops WHERE tier_two='$drop_var'")
or die(mysql_error());
echo '<select name="drop_3" id="drop_3">
<option value=" " disabled="disabled" selected="selected">Choose one</option>';
while($drop_3 = mysql_fetch_array( $result ))
{
echo '<option value="'.$drop_3['tier_three'].'">'.$drop_3['tier_three'].'</option>';
}
echo '</select> ';
echo '<input type="submit" name="submit" value="Submit" />';
}
?>
Gracias de antemano