php-general Digest 18 Jul 2011 22:01:16 -0000 Issue 7405
Topics (messages 314100 through 314102):
Re: chained select with ajax
314100 by: Tamara Temple
314101 by: Jim Lucas
How to sum monetary variables
314102 by: MartÃn Marqués
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----------------------------------------------------------------------
--- Begin Message ---
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.
--- End Message ---
--- Begin Message ---
On 7/17/2011 11:38 AM, Chris Stinemetz wrote:
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
I have made a few assumptions in the following code... with that said
Well, without seeing more of your code, from what I can tell from what
you provided, you are missing a few statements in your jQuery and you
are using the wrong variables that jQuery is passing to your PHP.
Also, since I do not see you populating the #market SELECT field with
jQuery, is it safe to assume that you are populating it when you create
the page?
> <script>
> $(document).ready(function()
> {
> $("select#type").attr("disabled","disabled");
> $("select#store").attr("disabled","disabled");
> $("select#market").change(function()
> {
remove the following line, it is redundant
> $("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()
> {
remove the following line, it is redundant
> $("select#store").attr("disabled","disabled");
> $("select#store").html("<option>please wait...</option>");
var m_id = $("select#market option:selected").attr('value');
var t_id = $("select#type option:selected").attr('value');
$.post("select_store.php",
{market_id:m_id,type_id:t_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("cannot 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))
> {
you are using $row['id'] below, but you do not select it above...
> $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))
> {
you are using $row['id'] below, but you do not select it above...
> $type .= '<option value="' . $row['id'] . '">' .
> $row['store_type'] . '</option>';
> }
> return $type;
> }
>
> public function ShowStore()
> {
> $sql = "SELECT store_name
> FROM store_list
Are you suppose to be using the same $_POST variable for this select
statement?
> WHERE id_markets=$_POST[id]
> AND store_type=$_POST[id]";
to work with the changes that I made above in the jQuery code, you will
need to change the previous two lines to the following two lines of code
> WHERE id_markets=$_POST['market_id']
> AND store_type=$_POST['type_id']";
That show fix most of your problems and get you headed down the rod to
recovery...
> $res = mysql_query($sql,$this->conn);
> $Store = '<option value="0">stores...</option>';
> while($row = mysql_fetch_array($res))
> {
you are using $row['id'] below, but you do not select it above...
> $Store .= '<option value="' . $row['id'] . '">' .
> $row['store_name'] . '</option>';
> }
> return $Store;
> }
> }
>
> $opt = new SelectList();
>
> ?>
--- End Message ---
--- Begin Message ---
I'm building a table (which is a report that has to be printed) with a
bunch of items (up to 300 in some cases) that have unitary price
(stored in a numeric(9,2) field), how many there are, and the total
price for each item. At the end of the table there is a total of all
the items.
The app is running on PHP and PostgreSQL is the backend.
The question is, how do I get the total of everything?
Running it on PHP gives one value, doing a sum() on the backend gives
another, and I'm starting to notice that even using python as a
calculator gives me errors (big ones). Right now I'm doing the maths
by hand to find out who has the biggest error, or if any is 100%
accurate.
Any ideas?
--
Martín Marqués
select 'martin.marques' || '@' || 'gmail.com'
DBA, Programador, Administrador
--- End Message ---
