On Jul 17, 2011, at 2:38 PM, Chris Stinemetz wrote:
This is a [Cross-post] I didn't receive any feedback from phpdb list.
Hope fully there is someone out there that may offer some advice why
my code isn't working correctly.
Thanks all.
I am trying to create a cascading seletct with 3 menu choices.
For some reason my third select menu is not populating. It doesn't
seem like my ajax is correctly sending $_post values to the final
query in my PHP class I built.
By the time I make it to the final third select menu there are no
choices.
Hopefully someone can find something I missed in the following code
snippits.
Any help is greatly appreciated.
Please excuse the incorrect indentions. For some reason gmail
changes it.
Thanks,
Chris
ajax code...
<script>
$(document).ready(function(){
$("select#type").attr("disabled","disabled");
$("select#store").attr("disabled","disabled");
$("select#market").change(function(){
$("select#type").attr("disabled","disabled");
$("select#type").html("<option>please wait...</option>");
var id = $("select#market option:selected").attr('value');
$.post("select_type.php", {id:id}, function(data){
$("select#type").removeAttr("disabled");
$("select#type").html(data);
});
});
$("select#type").change(function(){
$("select#store").attr("disabled","disabled");
$("select#store").html("<option>please
wait...</option>");
var id = $("select#type
option:selected").attr('value');
$.post("select_store.php", {id:id}, function(data){
$("select#store").removeAttr("disabled");
$("select#store").html(data);
});
});
$("form#select_form").submit(function(){
var market = $("select#market
option:selected").attr('value');
var type = $("select#type option:selected").attr('value');
var store = $("select#store
option:selected").attr('value');
if(market>0 && type>0 && store>0)
{
var market = $("select#market option:selected").html();
var type = $("select#type
option:selected").html();
var store = $("select#store
option:selected").html();
$("#result").html('your choices were: '+market +' ,
'+type +' and '+store);
}
else
{
$("#result").html("you must choose three options!");
}
return false;
});
});
</script>
php class for populating select menus....
<?php
class SelectList
{
protected $conn;
public function __construct()
{
$this->DbConnect();
}
protected function DbConnect()
{
include "db_config.php";
$this->conn =
mysql_connect($host,$user,$password) OR die("Unable
to connect to the database");
mysql_select_db($db,$this->conn) OR die("can
not select the database $db");
return TRUE;
}
public function ShowMarket()
{
$sql = "SELECT DISTINCT id_markets FROM
store_list";
$res = mysql_query($sql,$this->conn);
$market = '<option value="0">market...</
option>';
while($row = mysql_fetch_array($res))
{
$market .= '<option value="' .
$row['id'] . '">' .
$row['id_markets'] . '</option>';
}
return $market;
}
public function ShowType()
{
$sql = "SELECT DISTINCT store_type FROM
store_list";
$res = mysql_query($sql,$this->conn);
$type = '<option value="0">store type...</
option>';
while($row = mysql_fetch_array($res))
{
$type .= '<option value="' .
$row['id'] . '">' .
$row['store_type'] . '</option>';
}
return $type;
}
public function ShowStore()
{
$sql = "SELECT store_name FROM store_list WHERE
id_markets=$_POST[id] AND store_type=$_POST[id]";
$res = mysql_query($sql,$this->conn);
$Store = '<option value="0">stores...</option>';
while($row = mysql_fetch_array($res))
{
$Store .= '<option value="' .
$row['id'] . '">' .
$row['store_name'] . '</option>';
}
return $Store;
}
}
$opt = new SelectList();
?>
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It's a little hard to see what's happening here without the code for
select_type.php and select_store.php... You might try throwing in some
trace code sending the output to the error log (use error_log() php
function to do that) or dump the output you get from the ajax returns
to the client screen to see what is getting returned, if anything.
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