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>';
}
No comments posted yet
Your Answer:
Login to answer
286
15
Other forums
Problem executing bash script using shell_exec
Hi there,
I created a bash script file using following code to convert doc documents to pdf using
PHP using IF to display error
i have a MySQL query and i want to display 1 thing only if the number of affected rows is >=1
mysql_real_escape_string making variable equal nothing
i post a form and i post the variable:
Code: $var = $_POST[variable];then i echo $var its what i
Need Help with a query
Hello,
For some reason I am just not getting the right answer when I do this query and not quite
User information
Hi All,
By using t-code SUIM we get all the user details,if suppose we donot have permission t
Must be a string? Huh, what? HELP?
with the following script, I get
Fatal error: Property name must be a string in /home/content/e/
Creating a custom API
I'm creating a site, and I need to create a basic API. Unfortunately I have no idea where to start.
Can I use a loop
Hi buddies!
Once again with my doubts here.
Right now I am using this sql stat
help with mysql_error()
Hi,
I am trying to insert data into a table, but I am not able to insert it. I wanted to see
BIG file upload!
Hey guys!
I'm trying to upload a file, it works well with smaller files but with 60mb+, I get