sql get balance at end of year - sql

Question

I have a transactions table for a single year with the amount indicating the debit transaction if the value is negative or credit transaction values are positive.
Now in a given month if the number of debit records is less than 3 or if the sum of debits for a month is less than 100 then I want to charge a fee of 5.
I want to build and sql query for this in postgre:
select sum(amount), count(1), date_part('month', date) as month from transactions where amount < 0 group by month;
I am able get records per month level, I am stuck on how to proceed further and get the result.

Answer 1

You can start by generating the series of month with generate_series(). Then join that with an aggregate query on transactions, and finally implement the business logic in the outer query:
select sum(t.balance)
- 5 * count(*) filter(where coalesce(t.cnt, 0) < 3 or coalesce(t.debit, 0) < 100) as balance
from generate_series(date '2020-01-01', date '2020-12-01', '1 month') as d(dt)
left join (
select date_trunc('month', date) as dt, count(*) cnt, sum(amount) as balance,
sum(-amount) filter(where amount < 0) as debit
from transactions t
group by date_trunc('month', date)
) t on t.dt = d.dt
Demo on DB Fiddle:
| balance |
| ------: |
| 2746 |

Answer 2

How about this approach?
SELECT
SUM(
CASE
WHEN usage.amount_s > 100
OR usage.event_c > 3
THEN 0
ELSE 5
END
) AS YEAR_FEE
FROM (SELECT 1 AS month UNION
SELECT 2 UNION
SELECT 3 UNION
SELECT 4 UNION
SELECT 5 UNION
SELECT 6 UNION
SELECT 7 UNION
SELECT 8 UNION
SELECT 9 UNION
SELECT 10 UNION
SELECT 11 UNION
SELECT 12
) months
LEFT OUTER JOIN
(
SELECT
sum(amount) AS amount_s,
count(1) event_c,
date_part('month', date) AS month
FROM transactions
WHERE amount < 0
GROUP BY month
) usage ON months.month = usage.month;

Answer 3

First you must use a resultset that returns all the months (1-12) and join it with a LEFT join to your table.
Then aggregate to get the the sum of each month's amount and with conditional aggregation subtract 5 from the months that meet your conditions.
Finally use SUM() window function to sum the result of each month:
SELECT DISTINCT SUM(
COALESCE(SUM(t.Amount), 0) -
CASE
WHEN SUM((t.Amount < 0)::int) < 3
OR SUM(CASE WHEN t.Amount < 0 THEN -t.Amount ELSE 0 END) < 100 THEN 5
ELSE 0
END
) OVER () total
FROM generate_series(1, 12, 1) m(month) LEFT JOIN transactions t
ON m.month = date_part('month', t.date) AND date_part('year', t.date) = 2020
GROUP BY m.month
See the demo.
Results:
> | total |
> | ----: |
> | 2746 |

Answer 4

I think you can use the hanving clause.
Select ( sum(a.total) - (12- count(b.cnt ))*5 ) as result From
(Select sum(amount) as total , 'A' as name from transactions ) as a left join
(Select count(amount) as cnt , 'A' as name
From transactions
where amount <0
group by month(date)
having not(count(amount) <3 or sum(amount) >-100) ) as b
on a.name = b.name

Answer 5

select
sum(amount) - 5*(12-(
select count(*)
from(select month, count(amount),sum(amount)
from transactions
where amount<0
group by month
having Count(amount)>=3 And Sum(amount)<=-100))) as balance
from transactions ;