Problem with variable declaration in switch statement

Posted on 16th Feb 2014 07:03 pm by admin

Hello, I am having some trouble assigning a value to a variable inside a switch statement. What I am trying to do here is trip an error if the user has already added an item in the shopping cart. The While and Foreach loops work perfectly fine. The problem is that the value of variable $error does not get passed outside of the case. As a result, I cannot get the error message to display. I really need to get this to work and any help is appreciated. The code looks something like this:

Code: switch ($action) {
case 'add' :
{
if ($_SESSION['CartID'] == "")
{

$query = 'INSERT INTO ShopCart( UserID )VALUES("' . $UserID . '")';
mysql_query($query, $db) or die(mysql_error($db));

$Shop = 'SELECT ShopCartID FROM ShopCart WHERE UserID = "' . $UserID . '" ORDER BY ShopCartID ASC LIMIT 1';
$Cart = mysql_query($Shop);
$ShopCart = mysql_fetch_row($Cart);
$_SESSION['CartID'] = $ShopCart[0];
}
$query = 'SELECT ProductID FROM ShopCartLine WHERE ShopCartID = "' . $_SESSION['CartID'] . '"';
$result=mysql_query($query, $db) or die(mysql_error($db));

//$row = mysql_fetch_row( $result );
while($var = mysql_fetch_array($result))
{
foreach ($var as $i) {
if ($i == $ProductID) {
$error = 1; //TRIP THE ERROR HERE!
echo "<script type='text/javascript'>window.top.location='http://amarcy2.db-class.ids.uic.edu/shopcart.php';</script>";
}
}
}
}


The code being used to display the error message is as follows:

Code:
if ($error == 1) {
echo '<p>You already have this item in your basket!</p>';
}

Your Answer:

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