MySQL - Arithmetic operations with two relations - mysql

Question

How do I get the total amount of money paid by each customer minus the amount collected (em_paid_to)?
table customer
cust_id INT
f_name VARCHAR
l_name VARCHAR
email VARCHAR
c_limit INT
table transaction
id INT
em_paid_by VARCHAR
em_paid_to VARCHAR
amount INT
trans_date DATE
I already tried this to get the total paid by each customer but it did not work:
SELECT C.F_NAME, C.L_NAME, COUNT(T.EM_PAID_BY), SUM(T.AMOUNT)
FROM CUSTOMER C
JOIN TRANSACTION T ON C.EMAIL = T.EM_PAID_BY;
...and this to get the total collected by each customer, still the same error and I need to get the difference between the two results.
SELECT C.F_NAME, C.L_NAME, COUNT(T.EM_PAID_TO), SUM(T.AMOUNT)
FROM CUSTOMER C
JOIN TRANSACTION T ON C.EMAIL = T.EM_PAID_TO;
What I am hoping to get is like this Old McDonald oldmcdonald#gmail.com 2000
i.e (2000 + 4000 + 1000) - (2000 + 3000) = 2000

Answer 1

Your existing queries generate errors because they use aggregate functions (SUM(), COUNT()) without a GROUP BY clause to list all non-aggregated columns.
To solve your requirement, one solution would be to use conditional aggregation:
recover all transactions where the customer's email appears in the em_paid_to or in the em_paid_by clause
group by customer (a sensible option is to add the customer id to the GROUP BY clause, eventhough it is not part of the results)
do conditional counts and sums, depending on whether the records was matched on em_paid_to or em_paid_by
The following query gives you detailed information (count of payments by and to, amounts paid by and to, and balance), you can pick what's relevant for you:
SELECT
c.f_name,
c.l_name,
SUM(c.email = t.em_paid_by) count_paid_by,
SUM(c.email = t.em_paid_to) count_paid_to,
SUM(CASE WHEN c.email = t.em_paid_by THEN t.amound ELSE 0 END) total_paid_by,
SUM(CASE WHEN c.email = t.em_paid_to THEN t.amound ELSE 0 END) total_paid_to,
SUM(CASE WHEN c.email = t.em_paid_by THEN t.amound ELSE -1 * t.amount END) balance
FROM
customer c
INNER JOIN transaction t
ON c.email IN (t.em_paid_by, t.em_paid_to)
GROUP BY
c.cust_id,
c.f_name,
c.l_name
;

Answer 2

I would unpivot the data and aggregate:
select t.email, c.fname, c.lname, sum(t.amount)
from ((select em_paid_by as email, -amount as amount
from transaction t
) union all
(select em_paid_to, amount
from transaction t
)
) t
group by email;
You can join to the customer table to get the additional customer information:
select email, sum(amount)
from cusomer c join
((select em_paid_by as email, -amount as amount
from transaction t
) union all
(select em_paid_to, amount
from transaction t
)
) t
on c.email = t.email
group by t.email, c.fname, c.lname;

Answer 3

I would use correlated subqueries in the SELECT clause:
select c.*,
coalesce((
select sum(amount)
from transaction t
where t.em_paid_by = c.email
), 0)
-
coalesce((
select sum(amount)
from transaction t
where t.em_paid_to = c.email
), 0) as paid_balance
from customer c
If you want more information like the count of transactions, I would use subqueries in the FROM clause:
select c.*,
p.cnt_paid,
r.cnt_received
coalesce(p.sum_paid, 0) as sum_paid,
coalesce(r.sum_received, 0) as sum_received,
coalesce(p.sum_paid, 0) - coalesce(r.sum_received, 0) as paid_balance,
p.cnt_paid + r.cnt_received as total_transactions
from customer c
left join (
select em_paid_by as email, sum(amount) as sum_paid, count(*) as cnt_paid
from transaction
group by em_paid_by
) p on p.email = c.email
left join (
select em_paid_to as email, sum(amount) as sum_received, count(*) as cnt_received
from transaction
group by em_paid_to
) r on r.email = c.email