Question

我正在开发一款迷你购物车。一切似乎都有效，但是，当我点击其他产品时，它并没有提取正确的数据。

我选择产品1 - 它为我提供了正确的信息，但如果我选择产品2，它将提供与产品1相同的信息。

我将在下面显示我的代码，但我相信错误来自这行代码：

$sql = sprintf("SELECT Model, Price FROM Bike WHERE BikeCode = %d;", $bikecode);

我认为它没有获得$ bikecode

这是php文件中的整个代码：

<?php
    $bikecode = $_GET['id'];     //the product id from the URL 
    $action = $_GET['action']; //the action from the URL

if($bikecode && !productExists($bikecode)) {
    die("Product Doesn't Exist");
} 

    switch($action) {   //decide what to do 

        case "add":
            $_SESSION['cart'][$bikecode]++; //add one to the quantity of the product with id $bikecode 
        break;

        case "remove":
            $_SESSION['cart'][$bikecode]--; //remove one from the quantity of the product with id $bikecode 
            if($_SESSION['cart'][$bikecode] == 0) unset($_SESSION['cart'][$bikecode]); //if the quantity is zero, remove it completely (using the 'unset' function) - otherwise is will show zero, then -1, -2 etc when the user keeps removing items. 
        break;

        case "empty":
            unset($_SESSION['cart']); //unset the whole cart, i.e. empty the cart. 
        break;
    }

    if($_SESSION['cart']){

        echo "<table width=\"100%\">";   

          foreach($_SESSION['cart'] as $bikecode => $quantity) { 

              $sql = sprintf("SELECT Model, Price FROM Bike WHERE BikeCode = %d;", $bikecode); 

                $result = mysqli_query($con, $sql);

                if(mysqli_num_rows($result) > 0) {

                    list($model, $price) = mysqli_fetch_row($result);   

                    $cost = $quantity * $price;
                    $total = $total + $cost;

                        echo "<tr><th>Model:</th><th>Quantity:</th><th>Price:</th></tr>";
                        echo "<tr>";
                        echo "<td align=\"center\">$model</td>";
                        echo "<td align=\"center\">$quantity <a href=\"$_SERVER[PHP_SELF]?action=remove&id=$bikecode\">X</a></td>";
                        echo "<td align=\"center\">£$cost</td>";
                    echo "</tr>";
                }
            }

            echo "<tr>";
                echo "<td colspan=\"2\" align=\"right\">Total</td>";
                echo "<td align=\"right\">£$total</td>";
            echo "</tr>";

            echo "<tr>";
                echo "<td colspan=\"3\" align=\"right\"><a href=\"$_SERVER[PHP_SELF]?action=empty\" onclick=\"return confirm('Are you sure?');\">Empty Cart</a></td>";
            echo "</tr>";       
        echo "</table>";

    }else{
        echo "You have no items in your shopping cart.";
    }

function productExists($bikecode) {
    $sql = sprintf("SELECT * FROM Bike WHERE BikeCode = %d;", $bikecode); 
    return mysqli_num_rows(mysqli_query($con, $sql)) > 0;
}
?>

非常感谢帮助，但如果您找不到错误，请指导我以另一种方式显示此代码：

$sql = sprintf("SELECT Model, Price FROM Bike WHERE BikeCode = %d;", $bikecode);

Answer 1

看起来你已经得到了它，但是为了便于引用可能会搜索它的其他人，问题是你在sprintf（）中使用％d来尝试表示当％d被视为整数时的varchar。使用％s将解决问题。

sprintf() - Manual

迷你购物车有问题

1 个答案: