Ответов: 3 Просмотров: 1132
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Вот ajax кнопка добавления
CODE ( javascript):
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$('.add-cart-style-list,.add-cart-style-grid,.add-cart,.random-add-cart').click(function(){ var tid = $(this).attr("tid"); var size = $(this).attr("size"); $.ajax({ type: "POST", url: "/include/addtocart.php", data: "id="+tid+"&size"+size, dataType: "html", cache: false, success: function(data) { loadcart(); } });
Сама форма:
CODE ( html):
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<form method="POST" action="/include/addtocart.php" name="form"> <p>Размер: <select name="size" id="size" > <option value="41">41</option> <option value="42">42</option> <option value="43">43</option> <option value="44">44</option> <option value="45">45</option> </select> </p> </p> </form>
и обработчик:
PHP:
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<?PHP if($_SERVER["REQUEST_METHOD"] == "POST") { include("db_connect.php"); include("../functions/functions.php"); $id = clear_string($_POST["id"]); $result = mysql_query("SELECT * FROM cart WHERE cart_ip = '{$_SERVER['REMOTE_ADDR']}' AND cart_id_products = '$id'",$link); { $new_count = $row["cart_count"] + 1; $update = mysql_query ("UPDATE cart SET cart_count='$new_count' WHERE cart_ip = '{$_SERVER['REMOTE_ADDR']}' AND cart_id_products ='$id'",$link); } else { $result = mysql_query("SELECT * FROM tableproducts WHERE products_id = '$id'",$link); mysql_query("INSERT INTO cart(cart_id_products,cart_price,size,cart_datatime,cart_ip) VALUES( '".$row['products_id']."', '".$row['price']."', '".$_POST['size']."', NOW(), '".$_SERVER['REMOTE_ADDR']."' )",$link); } } ?>
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