[SOLVED] PHP/MySQL Results Invalid

[techtheatre]

I know this question probably has more to do with SQL than PHP, except i have checked the SQL and it seems fine.  I have a basic shopping cart script i am building and the items in the cart are stored in a table in my DB.  I have the following function that should display a "quick summary" of the number of items in the cart.  Earlier on the page i have included { really i used a require_once() } the $db connection info.

 

function showCartSummary() {
$NumItemsInCart = "0";  //reset to zero for default
$sql = "SELECT SUM(Quantity) FROM ".DB_TABLE_PREFIX."CartItems WHERE CartId='$cartid' ";
$result = @mysql_query($sql, $db);
if($result)
{
	$row = mysql_fetch_array($result);
	$NumItemsInCart = $row[0];
}

if ($NumItemsInCart == "0") {
	return "$sql<br />$NumItemsInCart<p><font size=\"-1\">You have no items in your shopping cart</font></p>";
} else {
	$s = ($NumItemsInCart > 1) ? 's':'';  //make item(s) plural if more than one
	return "$sql<br />$NumItemsInCart<p><font size=\"-1\"><a href=\"cart.php\">You have $NumItemsInCart item".$s." <br />in your cart</a></font><br /><a href=\"cart.php\">VIEW CART</a></p>";
}
}

I am having the query returned at this point for troubleshooting, and get the correct query returned as follows:

SELECT SUM(Quantity) FROM demo_CartItems WHERE CartId='123'

 

When i copy/paste that query into phpMyAdmin it properly returns a value of "9"...but when i run the function it always says i have no items in my cart (first option).

 

Here is the data in my table (demo_CartItems):

 

Item CartId ItemName ItemId Quantity

3 123 TST 251 4

2 123 Product2 250 1

8 123 CPY 252 2

7 123 CPY 252 1

9 123 CPY 252 1

 

 

Any help is greatly appreciated.  I have been hitting my head on this one for a while now...i am sure i have stupidly overlooked something obvious.  Thanks.

 

[Glyde]

function showCartSummary() {
        global $db;
$NumItemsInCart = "0";  //reset to zero for default
$sql = "SELECT SUM(Quantity) AS qty FROM ".DB_TABLE_PREFIX."CartItems WHERE CartId='$cartid' ";
$result = @mysql_query($sql, $db);
if($result)
{
	$row = mysql_fetch_array($result);
	$NumItemsInCart = $row['qty'];
}

if (!$NumItemsInCart) {
	return "$sql<br />$NumItemsInCart<p><font size=\"-1\">You have no items in your shopping cart</font></p>";
} else {
	$s = ($NumItemsInCart > 1) ? 's':'';  //make item(s) plural if more than one
	return "$sql<br />$NumItemsInCart<p><font size=\"-1\"><a href=\"cart.php\">You have $NumItemsInCart item".$s." <br />in your cart</a></font><br /><a href=\"cart.php\">VIEW CART</a></p>";
}
}

 

The "AS" syntax should be what you need here.  Look how I modified the query and the result returned so you can see for future reference.

[techtheatre]

i had already tried it with an AS line in there (which i do not commonly use).  Unfortunately that still gave the same result.

 

You DID however solve the problem...but i think it was the inclusion of the following:

 

        global $db;

 

I will have to read a bit more to find out exactly what that does, but it seems to have solved the problem! 

THANKS!

[Glyde]

PHP, as almost every other programming language (I believe), has variable scopes.  This means that variables defined outside of a function and a class are on a different scope than those defined in a class, and again in a different scope from those defined in a function.  If you want to access a variable that was declared outside of a function, meaning it was defined globally, you must precede the variable with global, which tells PHP to pull that variable from the global scope, not the internal function scope.