Group By - select by a criteria that is met every month - sql

Question

The below query returns all USERS that have SUM(AMOUNT) > 10 in a given month. It includes Users in a month even if they don't meet the criteria in other months.
But I'd like to transform this query to return all USERS who must meet the criteria SUM(AMOUNT) > 10 every single month (i.e., from the first month in the table to the last one) across the entire data.
Put another way, exclude users who don't meet SUM(AMOUNT) > 10 every single month.
select USERS, to_char(transaction_date, 'YYYY-MM') as month
from Table
GROUP BY USERS, month
HAVING SUM(AMOUNT) > 10;

Answer 1

One approach uses a generated calendar table representing all months in your data set. We can left join this calendar table to your current query, and then aggregate over all months by user:
WITH months AS (
SELECT DISTINCT TO_CHAR(transaction_date, 'YYYY-MM') AS month
FROM yourTable
),
cte AS (
SELECT USERS, TO_CHAR(transaction_date, 'YYYY-MM') AS month
FROM yourTable
GROUP BY USERS, month
HAVING SUM(AMOUNT) > 10
)
SELECT
t.USERS
FROM months m
LEFT JOIN cte t
ON m.month = t.month
GROUP BY
t.USERS
HAVING
COUNT(t.USERS) = (SELECT COUNT(*) FROM months);
The HAVING clause above asserts that the number of months to which a user matches is in fact the total number of months. This would imply that the user meets the sum criteria for every month.

Answer 2

Perhaps you could use a correlated subquery, such as:
select t.*
from (select distinct table.users from table) t
where not exists
(
select to_char(u.transaction_date, 'YYYY-MM') as month
from table u
where u.users = t.users
group by month
having sum(u.amount) <= 10
)

Answer 3

One option would be using sign(amount-10) vs. sign(amount) logic as
SELECT q.users
FROM
(
with tab(users, transaction_date,amount) as
(
select 1,date'2018-11-24',8 union all
select 1,date'2018-11-24',18 union all
select 2,date'2018-10-24',13 union all
select 3,date'2018-11-24',18 union all
select 3,date'2018-10-24',28 union all
select 3,date'2018-09-24', 3 union all
select 4,date'2018-10-24',28
)
SELECT users, to_char(transaction_date, 'YYYY-MM') as month,
sum(sign(amount-10)) as cnt1,
sum(sign(amount)) as cnt2
FROM tab t
GROUP BY users, month
) q
GROUP BY q.users
HAVING sum(q.cnt1) = sum(q.cnt2)
GROUP BY q.users
users
-----
2
4
Rextester Demo

Answer 4

You need to compare the number of months > 10 to the number of months between the min and the max date:
SELECT users, Count(flag) AS months, Min(mth), Max(mth)
FROM
(
SELECT users, date_trunc('month',transaction_date) AS mth,
CASE WHEN Sum(amount) > 10 THEN 1 end AS flag
FROM tab t
GROUP BY users, mth
) AS dt
GROUP BY users
HAVING -- adding the number of months > 10 to the min date and compare to max
Min(mth) + (INTERVAL '1' MONTH * (Count(flag)-1)) = Max(mth)
If missing months don't count it would be a simple count(flag) = count(*)