MySQL SUM of one column, DISTINCT of ID column - mysql

Question

I'm trying to create a summary report of our orders but having trouble extracting all my required data in a single query.
The data I'd like to extract:
subtotal - SUM of all sale prices
delivery total - SUM of all orders deliveryTotal
orders - COUNT of DISTINCT orderIds
quantity - SUM of all quantity ordered
Orders table (simplified for this example)
| orderId | deliveryTotal | total |
|---------|---------------|-------|
| 1 | 5 | 15 |
| 2 | 5 | 15 |
| 3 | 7.50 | 27.50 |
Order items table
| orderItemId | orderId | productId | salePrice | quantity |
|-------------|---------|-----------|-----------|----------|
| 1 | 1 | 1 | 10 | 1 |
| 2 | 2 | 1 | 10 | 1 |
| 3 | 3 | 1 | 10 | 1 |
| 4 | 3 | 2 | 10 | 1 |
My current query for extracting this data is
SELECT
SUM(i.salePrice * i.quantity) as subtotal,
SUM(DISTINCT o.deliveryTotal) as deliveryTotal,
COUNT(DISTINCT o.orderId) as orders,
SUM(i.quantity) as quantity
FROM orderItems i
INNER JOIN orders o ON o.orderId = i.orderId
Which results in a correct subtotal, order count and quantity sum. But delivery total is returned as 12.50 when I'm after 17.50. If I do SUM(o.deliveryTotal) it will return 25.
EDIT: Desired results
| subtotal | deliveryTotal | orders | quantity |
|----------|---------------|--------|----------|
| 40.00 | 17.50 | 3 | 4 |

Answer 1

https://tiaashish.wordpress.com/2014/01/31/mysql-sum-for-distinct-rows-with-left-join/
Here is a blog post that shows exactly what I was looking for. Maybe this can help others too.
The formula is something like this:
SUM(o.deliveryTotal) * COUNT(DISTINCT o.orderId) / COUNT(*)

Answer 2

Because of the join, the SUM(DISTINCT deliveryTotal) aggregate is being applied to a rowset including the values 5, 5, 7.5, 7.5 (distinct 5 + 7.5 = 12.5).
The rows your SUM() acted on become more apparent if you simply do
SELECT o.*
FROM orderItems i
INNER JOIN orders o ON o.orderId = i.orderId
Instead you are asking for the SUM() of all the values in deliveryTotal, irrespective of their position in the join with orderItems. That means you need to apply the aggregate at a different level.
Since you are not intending to add a GROUP BY later, the easiest way to do that is to use a subselect whose purpose is only to get the SUM() across the whole table.
SELECT
SUM(i.salePrice * i.quantity) as subtotal,
-- deliveryTotal sum as a subselect
(SELECT SUM(deliveryTotal) FROM orders) as deliveryTotal,
COUNT(DISTINCT o.orderId) as orders,
SUM(i.quantity) as quantity
FROM orderItems i
INNER JOIN orders o ON o.orderId = i.orderId
Subselects are usually discouraged but there won't be a significant performance penalty for the subselect, none different from the alternative methods of using a join for it. The calculation has to be done on a separate aggregate from the existing join no matter what. Other methods would place a subquery CROSS JOIN in the FROM clause, which performs the same thing we placed here in the subselect. Performance would be the same.

Answer 3

Select per Order in the Inner Select and than sum it up
Select
SUM(subtotal) as subtotal,
sum(deliveryTotal) as deliveryTotal,
count(1) as orders,
sum(quantity) as quantity
from (
SELECT
SUM(i.salePrice * i.quantity) as subtotal,
o.deliveryTotal as deliveryTotal,
SUM(i.quantity) as quantity
FROM orders o
INNER JOIN orderItems i ON o.orderId = i.orderId
group by o.orderId) as sub

Answer 4

The below query results exactly what you need
SELECT SUM(conctable.subtotal),
SUM(conctable.deliveryTotal),
SUM(conctable.orders),
SUM(conctable.quantity) from
(SELECT SUM(i.salePrice * i.quantity) as subtotal,
o.deliveryTotal as deliveryTotal,
COUNT(DISTINCT o.orderId) as orders,
SUM(i.quantity) as quantity
FROM orderItems i
JOIN orders o ON o.orderId = i.orderId group by i.orderid) as conctable;