选择两个不同表的总和

Question

I have an orders table which consists of the following:

id
order_total
delivery_cost
customer_id

I also have a transactions table which has:

id
amount
customer_id
status

What I'm trying to do is,

SELECT SUM(order_total + delivery_cost) FROM orders WHERE customer_id = '1'

then

SELECT SUM(amount) FROM transactions WHERE customer_id = '1' AND transaction_status = 'Paid'

Then with the data, minus the total amount from the order totals.

I've tried different queries using JOINS, but I just can't get my head around it, for example:

SELECT SUM(OrdersTotal - TransactionTotal) as AccountBalance

FROM (
SELECT SUM(`order_total` + `delivery_cost`) FROM `orders` as OrdersTotal 
UNION ALL 
SELECT SUM(`amount`) FROM `transactions` WHERE `transaction_status` = 'Paid' as TransactionTotal
) 

but this didn't work at all. Any help would be greatly appreciated.

Answer 1

The two datasets are effectively autonomous but assuming that it's unlikely to have transactions for a customer without orders, you can acheive your result with a LEFT JOIN rather than a ful outer join - but if you simply join the base tables then you'll likely get values from one table repeated in the interim result set (this is why Joseph B's answer is wrong when a customer has something other than a single row in each table).

SELECT ordered_value-IFNULL(paid_value,0) AS acct_balance
FROM
(
  SELECT customer_id, SUM(order_total + delivery_cost) AS ordered_value
  FROM orders 
  WHERE customer_id = '1'
  GROUP BY customer_id
) AS orders
LEFT JOIN 
(
  SELECT customer_id, SUM(amount) AS paid_value
  FROM transactions 
  WHERE customer_id = '1' 
  AND transaction_status = 'Paid'
  FROUP BY customer_id
) as payments
ON orders.customer_id = payments.customer_id

(here the 'GROUP BY' and 'ON' clauses are redundant since your only looking at a single customer - but are required for multiple customers).

Note that calculating a balance based on a sum of transactions is technically correct, it does not scale well - for large systems a better solution (although it breaks the rules of normalization) is to maintain a unified table of transactions and a balance for for the account along with each transaction amount - alternatively use checkpointing.

Answer 2

You can just name the column in your union all query and sum on that:

SELECT SUM(col) as AccountBalance
FROM (SELECT SUM(`order_total` + `delivery_cost`) as col FROM `orders` as OrdersTotal 
      UNION ALL 
      SELECT SUM(`amount`) FROM `transactions` WHERE `transaction_status` = 'Paid' as    TransactionTotal
     ) t;

Answer 3

Try this query using a JOIN:

SELECT
    SUM(o.order_total + o.delivery_cost) - SUM(t.amount) AS AccountBalance
FROM orders o
INNER JOIN transactions t
ON o.customer_id = t.customer_id AND o.customer_id = '1' AND t.transaction_status = 'Paid';

Answer 4

This should give you the account balance for each customer?

SELECT 
    ISNULL(o.customer_id, t.customer_id) AS customer_id
    OrdersTotal - TransactionTotal AS AccountBalance
FROM (
    SELECT 
        customer_id, 
        SUM(order_total + delivery_cost) AS OrdersTotal 
    FROM 
        orders 
    GROUP BY 
        customer_id) o
FULL OUTER JOIN (
    SELECT 
        customer_id, 
        SUM(amount) AS TransactionTotal 
    FROM 
        transactions 
    WHERE 
        transaction_status = 'Paid' 
    GROUP BY 
        customer_id) t
ON t.customer_id = o.customer_id

Answer 5

You can join the results of your queries and perform your calculations ,note sum without group by will result as one row so the sum in outer query doesn't mean when you have only one row and according to your logic of calculation union has nothing to do with it

SELECT t1.OrdersTotal - t2 .TransactionTotal AS AccountBalance
FROM (
SELECT SUM(order_total + delivery_cost) OrdersTotal ,customer_id 
 FROM orders 
 WHERE customer_id = '1'
) t1
JOIN (
SELECT SUM(amount) TransactionTotal ,customer_id
FROM transactions 
WHERE customer_id = '1' AND transaction_status = 'Paid'
) t2 USING(customer_id)

Answer 6

Since you are only concerned about a single customer, just have them listed as two different queries as the source...

select
      charges.chg - paid.pay as Balance
   from
      ( SELECT SUM(order_total + delivery_cost) chg
           FROM orders 
           WHERE customer_id = '1' ) charges,
      ( SELECT SUM(amount)  pay
           FROM transactions 
           WHERE customer_id = '1' AND transaction_status = 'Paid' ) paid

Now, if you wanted for all customers to see who is outstanding, add the customer's ID to each query and apply a group by, but then change to a LEFT-JOIN so you get all orders with or

select
      charges.customer_id,
      charges.chg - coalesce( paid.pay, 0 ) as Balance
   from
      ( SELECT customer_id, SUM(order_total + delivery_cost) chg
           FROM orders 
           group by customer_id ) charges
         LEFT JOIN ( SELECT customer_id, SUM(amount)  pay
                        FROM transactions 
                        where transaction_status = 'Paid' 
                        group by customer_id ) paid
          on charges.customer_id = paid.customer_id

Answer 7

I think this works best if you try an inline view. In the code block below, check the Group By clause, you might need to add more fields in the Group By depending on what you're selecting in the Inner SELECT statements.Try the code below:

SELECT
  SUM(Totals.OrdersTotal-Totals.TransactionTotal)
  FROM 
      (SELECT
           SUM(ord.order_total + ord.delivery_cost) AS OrdersTotal
           , SUM(trans.amount) AS TransactionTotal
           FROM orders ord
                INNER JOIN transactions trans
                     ON ord.customer_id = trans.customer_id
           WHERE
                ord.customer_id =1
                AND trans.transaction_status = 'Paid'
           GROUP BY
                ord.customer_id
      ) Totals;