SELECT with LEFT JOIN performing math operation twice? - sql

Question

The general idea of what I'm trying to do is this:
Select all planned prices for an order, then subtract from that total all actual prices on that order.
The planned price and actual price are on different tables. When I have a single planned price and a single actual price, this works fine. However, when I have multiple planned prices or multiple actual prices it is giving me odd results as if the algebra is happening multiple times.
Query:
SELECT PL.orderid, (SUM(PL.lineprice) - NVL(SUM(AC.lineprice),0)) AS
Difference FROM plans PL
LEFT JOIN actuals AC ON PL.orderid = AC.orderid
WHERE PL.customer IN (SELECT customer FROM ...)
GROUP BY PL.orderid
ORDER BY PL.orderid;
The results of the query:
Orderid Difference
X-1224 100
X-1226 80
X-1345 70000
X-1351 125000
X-1352 10000
Y-2403 190000
My Plan table looks like this:
Orderid Planned_Price
X-1224 100
X-1226 100
X-1345 105000
X-1351 100000
X-1352 10000
X-1352 50000
Y-2403 25000
Y-2403 100000
And my Actual table this:
Orderid Actual_Price
X-1226 20
X-1345 35000
X-1351 25000
X-1351 50000
X-1352 25000
Y-2403 25000
Y-2403 5000
So it seems to work when I have only a single row in each table, or a single row in plans and no rows in actuals i.e., X-1224, X-1226 and X-1345.
However the results are too high or too low when I have multiple rows, with the same OrderID, in either table i.e., all the rest
I'm stumped as to why this is the case. Any insights are appreciated.
edit: Results I'd like, taking Y-2403 as example: (25000 + 100000) - (25000 + 5000) = 95000. What I'm getting is double that at 190000.

Answer 1

Why is this the case?
Because that is how join works. If you have data like this:
a
1
1
2
2
And b:
b
1
1
1
2
Then the result of a join will have six "1"s and two "2"s.
Your question doesn't say what you want for results, but a typical approach is to aggregate before doing the joins.
EDIT:
You seem to want:
select p.orderid,
(p.lineprice - coalesce(lineprice, 0)) as Difference
from (select orderid, sum(lineprice) as lineprice
from plans p
group by orderid
) p left join
(select orderid, sum(lineprice) as lineprice
from actuals a
group by orderid
) a
on p.orderid = a.orderid
where p.customer in (SELECT customer FROM ...)
order by p.orderid;

Answer 2

I suppose you are looking to compare the summed_up_prices by order id of plan table with the summed_up prices by order id actual plan table.?
If so the following can be done to ensure there are no duplicates entries by order
select a.orderid
,NVL(max(b.summed_up),0) - sum(a.actual_price) as difference
from actual_table a
left join (select pt.orderid
,sum(pt.planned_price) as summed_up
from planned_table pt
group by pt.orderid
)b
on a.orderid=b.orderid
group by a.orderid
+---------+------------+
| ORDERID | DIFFERENCE |
+---------+------------+
| X-1226 | 80 |
| Y-2403 | 95000 |
| X-1351 | 25000 |
| X-1345 | 70000 |
| X-1352 | 35000 |
+---------+------------+
Here is the dbfiddle link with the data
https://dbfiddle.uk/?rdbms=oracle_11.2&fiddle=3cacffd19b39ecaf7ad752dff262ac47