-2

I want to insert data from input form and at the same time from another table (shopping_cart) in one table (order).

I found that there is no problem with passing value from input form or select it from shopping cart, since I print all the value that I want to use to insert in table (order) and it displayed.

But, when execute a query to insert into table order, the output is fail and nothing insert in database.

Below is my code :

<?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++;

}

?>
H.S
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1 Answers1

-1

Try this

$query = mysql_query("INSERT INTO order (ORDER_TOTALPRICE, ITEM_NAME, 
    CART_PRICE, CART_QUANTITY, FULLNAME, ADDRESS, PHONE) VALUES (".$total_price.",".$item_name.",".$cart_price.",".$cart_quantity.",".$fullname.",".$address.",".$phone.")");
  • Still the same, the output is 'fail'. Actually, I did the INSERT query same way as another operation/process before this, it worked. But I dont know why it's not work this time. – H.S Dec 17 '17 at 06:13