lukelee Posted January 28, 2009 Share Posted January 28, 2009 Hi, guys, I am making a simple shopping cart, now i got problem on check out function. i want to save the product information into database after click "checkout". because people may have different products in their shopping cart, so i made an array to display the items. but when I POST those items to next step which is check out, it only posts the latest item's information. can anyone help? thanks, here is a part of my code: function checkout() { $output[] = '<form action="checkout_process.php" method="post" id="cart">'; global $db; $cart = $_SESSION['cart']; if ($cart) { $items = explode(',',$cart); $contents = array(); foreach ($items as $item) { $contents[$item] = (isset($contents[$item])) ? $contents[$item] + 1 : 1; } $output[] = '<table>'; foreach ($contents as $id=>$qty) { $sql = 'SELECT * FROM product WHERE product_id = '.$id; $result = $db->query($sql); $row = $result->fetch(); extract($row); $output[] = '<tr><td><input type="hidden" name="model" value="'.$model.'"/></td>'; $output[] = '<td><input type="hidden" name="quantity'.$id.'" value="'.$qty.'" size="3" maxlength="3" /></td>'; $output[] = '<td><input type="hidden" name="price" value="'.$price.'"/></td>'; $output[] = '<td><input type="hidden" name="price" value="'.($price * $qty).'"/></td>'; $total += $price * $qty; $output[] = '</tr>'; } $output[] = '</table>'; $output[] = '<div><button type="submit">check out</button></div>'; $output[] = '</form>'; } else { $output[] = '<p>You shopping cart is empty.</p>'; } return join('',$output); } Link to comment https://forums.phpfreaks.com/topic/142779-need-help-on-shoping-cart-check-out/ Share on other sites More sharing options...
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