我想从输入表单中插入数据,同时从另一个表(shopping_cart)中插入一个表(顺序)。
我发现从输入表单传递值或从购物车中选择它没有问题,因为我打印了我想要用于在表(顺序)中插入的所有值并显示它。
但是,当执行要插入表顺序的查询时,输出失败并且没有任何内容插入数据库。
以下是我的代码:
<?php
include('dbconnection.php');
$query1 = mysql_query("SELECT * FROM shopping_cart");
$bil = 1;
while ($data = mysql_fetch_array($query1)) {
//FROM TABLE SHOPPING_CART
$item_name=$data['ITEM_NAME'];
$cart_price=$data['CART_PRICE'];
$cart_quantity=$data['CART_QUANTITY'];
//FROM HTML FORM
$total_price = $_POST['ORDER_TOTALPRICE'];
$fullname = mysql_real_escape_string($_POST['FULLNAME']);
$address = mysql_real_escape_string($_POST['ADDRESS']);
$phone = mysql_real_escape_string($_POST['PHONE']);
echo $item_name;
echo $cart_price;
echo $cart_quantity;
echo $fullname;
echo $address;
echo $phone;
echo $total_price;
//SOMETHING WRONG HERE ?
$query = mysql_query("INSERT INTO order (ORDER_TOTALPRICE, ITEM_NAME,
CART_PRICE, CART_QUANTITY, FULLNAME, ADDRESS, PHONE) VALUES ('$total_price','$item_name','$cart_price','$cart_quantity','$fullname','$address','$phone')");
if ($query){
echo "success";
}
else{
echo "fail";
}
$bil++;
}
?>
答案 0 :(得分:-1)
试试这个
$query = mysql_query("INSERT INTO order (ORDER_TOTALPRICE, ITEM_NAME,
CART_PRICE, CART_QUANTITY, FULLNAME, ADDRESS, PHONE) VALUES (".$total_price.",".$item_name.",".$cart_price.",".$cart_quantity.",".$fullname.",".$address.",".$phone.")");