Shopping cart for multiple tables

[ashraf02]

basicaly i am creating a shopping cart for my website however creating a shopping cart for one table of items is easy but can someone direct me on how to create a shopping cart that can handle more than one table of items. i really need ur help on this

thanks

ashraf

[ashraf02]

to elaborate further on my problem i am trying to execute the code below there are two seperate item tables. 1, shirts 2, trousers and i am trying to integrate it so that when customers select the item it would appear on the shopping basket even though there arein seperate tables. hope this makes sense.

Code:

[PHP]<?php
session_start() ;

$conn = mysql_connect(" localhost", "root", "")
or die (mysql_error()) ;
mysql_select_db ("noble", $conn) or die (mysql_error()) ;

if ($_POST[Bas_Item_ID] != "") {
$shirt_info = "SELECT Shirt_Brand
FROM Shirts
WHERE Shirt_ID =$_POST[Bas_Item_ID]";
$shirt_info_res = mysql_query ($shirt_info) or die (mysql_error()) ;

if (mysql_num_rows ($shirt_info_re s) < 1) {
header ("Location: Display_Smart.p hp");
exit;
} else {
$trouser_info = "SELECT Trouser_Brand
FROM Trousers
WHERE Trouser_ID =$_POST[Bas_Item_ID]";
$trouser_info_r es = mysql_query ($trouser_info) or die (mysql_error()) ;

if (mysql_num_rows ($trouser_info_ res) < 1) {
header ("Location: Display_Smart.p hp");
exit;
} else {
$trouser_brand = mysql_result($t rouser_info_res ,0,'Trouser_Bra nd');
$shirt_brand = mysql_result($s hirt_info_res,0 ,'Shirt_Brand') ;
}
$addcart = "INSERT INTO Basket
VALUES ('','$[COOKIE]PHPSESSID', '$_POST[Bas_Item_ID]', '$_POST[Bas_Item_Qty]', '$_POST[Bas_Item_Size]',
now())";

mysql_query($ad dcart);

header("locatio n: Showcart.php");
exit;

}

} else {
header ("Location: Display_Smart.p hp");
exit;
}

?>[/PHP]

thanks in advance

[ashraf02]

ive updated the code so that the shirts can go through to the basket but when i try putting the trousers through the basket nothng happens. please please please help it's urgent.

ashraf

updated code[php]
<?php
session_start() ;
$conn = mysql_connect(" localhost", "root", "")
or die (mysql_error()) ;
mysql_select_db ("noble", $conn) or die (mysql_error()) ;
if ($_POST[Bas_Item_ID] != "") {
$shirt_info = "SELECT Shirt_Brand
FROM Shirts
WHERE Shirt_ID =$_POST[Bas_Item_ID]";
$shirt_info_res = mysql_query ($shirt_info) or die (mysql_error()) ;

if (mysql_num_rows ($shirt_info_re s) < 1) {
header ("Location: Display_Smart.p hp");
exit;
} else {
$shirt_brand = mysql_result($s hirt_info_res,0 ,'Shirt_Brand') ;
$addcart = "INSERT INTO Basket
VALUES ('','$[COOKIE]PHPSESSID', '$_POST[Bas_Item_ID]', '$_POST[Bas_Item_Qty]', '$_POST[Bas_Item_Size]',
now())";
mysql_query($ad dcart);
header("locatio n: Showcart.php");
exit;
}
if ($_POST[Bas_Item_ID] != "") {
$trouser_info = "SELECT Trouser_Brand
FROM Trousers
WHERE Trouser_ID = $_POST[Bas_Item_ID]";
$trouser_info_r es = mysql_query ($trouser_info) or die (mysql_error()) ;
if (mysql_num_rows ($trouser_info_ res) < 1) {
header ("Location: Display_Smart.p hp");
exit;
} else {
$trouser_brand = mysql_result($t rouser_info_res ,0,'Trouser_Bra nd');

$addcart = "INSERT INTO Basket
VALUES ('','$[COOKIE]PHPSESSID', '$_POST[Bas_Item_ID]', '$_POST[Bas_Item_Qty]', '$_POST[Bas_Item_Size]',
now())";
mysql_query($ad dcart);
header("locatio n: Showcart.php");
exit;
}
} else {
header ("Location: Display_Smart.p hp");
exit;
}
}
?>[/php]

[ronverdonk]

As a full member now, you should know that we expect your code to be posted in [CODE] tags (See How to Ask a Question).

This makes it easier for our Experts to read and understand it. Failing to do so creates extra work for the moderators, thus wasting resources, otherwise available to answer the members' questions.

Please use the tags in future.

MODERATOR.

[ashraf02]

i apologise it was a mistake. any chance u could help me with my code though i am gettin really frustrated and dont know wat to do.

[TheServant]

I am not following what the problem is... Also, I am confused why you only have one POST variable (Bas_Item_ID). Both if statements are the same:
6. if ($_POST[Bas_Item_ID] != "") {
24. if ($_POST[Bas_Item_ID] != "") {

What error are you getting? What exactly is coming up, or not coming up?

[ashraf02]

for the the shirt table i can go through to the checkout basket but for the trousers table it comes up with a blank screen and i want both items to be added to the cart but only one table allows it. is there anyway i can add extra code to allow trousers to go through to the cart.

[code green]

You have a blank screen because you have no error checking or echo statements.
I have looked at the code and it is quite obvious what is happening -
If you don't buy a shirt the script exits.
As TheServant pointed out, one POST variable is used for both shirts and trousers.
Now please put some error checking and echo statements in your code

[ashraf02]

thanks for the response i have updated the code which looks like this :

[PHP]
<?php

session_start() ;

$conn = mysql_connect(" localhost", "root", "")

or die (mysql_error()) ;

mysql_select_db ("noble", $conn) or die (mysql_error()) ;

if ($_POST[Bas_Item_ID] != "") {

$shirt_info = "SELECT Shirt_Brand

FROM Shirts

WHERE Shirt_ID =$_POST[Bas_Item_ID]";

$shirt_info_res = mysql_query ($shirt_info) or die (mysql_error()) ;
if (mysql_num_rows ($shirt_info_re s) < 1) {
die (mysql_error()) ;
exit;
} else {
$shirt_brand = mysql_result($s hirt_info_res,0 ,'Shirt_Brand') ;
}
$trouser_info = "SELECT Trouser_Brand
FROM Trousers
WHERE Trouser_ID = $_POST[Bas_Item_ID]";
$trouser_info_r es = mysql_query ($trouser_info) or die (mysql_error()) ;
if (mysql_num_rows ($trouser_info_ res) < 1) {
die (mysql_error()) ;
exit;
} else {
$trouser_brand = mysql_result($t rouser_info_res ,0,'Trouser_Bra nd');

$addcart = "INSERT INTO Basket
VALUES ('','$[COOKIE]PHPSESSID', '$_POST[Bas_Item_ID]', '$_POST[Bas_Item_Qty]', '$_POST[Bas_Item_Size]',
now())";
mysql_query($ad dcart);
header("locatio n: Showcart.php");
exit;
}
} else {
header ("Location: Display_Smart.p hp");
exit;
}
?>[/PHP]

the error i get is :

Warning: session_start() [function.sessio n-start]: Cannot send session cache limiter - headers already sent (output started at C:\wamp\www\add tocart.php:2) in C:\wamp\www\add tocart.php on line 4

Warning: Cannot modify header information - headers already sent by (output started at C:\wamp\www\add tocart.php:2) in C:\wamp\www\add tocart.php on line 45

[code green]

This means you are trying to send header information to the browser
when output has already been sent or header information has already been sent.

[ashraf02]

how can i sort this problem out

[code green]

how can i sort this problem out

Well, all the information is in the error message.
On which line the problem occurred and in which file.
You cannot send multiple header information to a browser