create additional date after and before current row and create new column based on it

Question

Lets say I have this kind of data

create table example
(cust_id VARCHAR, product VARCHAR, price float, datetime varchar);

insert into example (cust_id, product, price, datetime)
VALUES
('1', 'scooter', 2000, '2022-01-10'),
('1', 'skateboard', 1500, '2022-01-20'),
('1', 'beefmeat', 300, '2022-06-08'),
('2', 'wallet', 200, '2022-02-25'),
('2', 'hairdryer', 250, '2022-04-28'),
('3', 'skateboard', 1600, '2022-03-29')

I want to make some kind of additional rows, and after that make new column based on this additional rows

My expectation output will like this

cust_id	total_price	date	is_active
1	3500	2022-01	active
1	0	2022-02	active
1	0	2022-03	active
1	0	2022-04	inactive
1	0	2022-05	inactive
1	300	2022-06	active
1	0	2022-07	active
2	0	2022-01	inactive
2	200	2022-02	active
2	0	2022-03	active
2	250	2022-04	active
2	0	2022-05	active
2	0	2022-06	active
2	0	2022-07	inactive
3	0	2022-01	inactive
3	0	2022-02	inactive
3	1600	2022-03	active
3	0	2022-04	active
3	0	2022-05	active
3	0	2022-06	inactive
3	0	2022-07	inactive

the rules is like this

1. the first month when the customer make transaction is called active, before this transaction called inactive.
   ex: first transaction in month 2, then month 2 is active, month 1 is inactive (look cust_id 2 and 3)
2. if more than 2 months there isnt transaction, the next month is inactive until there is new transaction is active.
   ex: if last transaction in month 1, then month 2 and month 3 is inactive, and month 4, month 5 inactive if month 6 there is new transaction (look cust_id 1 and 3)

well my first thought is used this code, but I dont know what the next step after it

select *, 
    date_part('month', age(to_date(date, 'YYYY-MM'), to_date(lag(date) over (partition by cust_id order by date),'YYYY-MM')))date_diff
from(
    select 
        cust_id,
        sum(price)total_price,
        to_char(to_date(datetime, 'YYYY-MM-DD'),'YYYY-MM')date
    from example
    group BY
        cust_id,
        date
    order by 
    cust_id, 
    date)test

I'm open to any suggestion

Answer 1

Try the following, an explanation within query comments:

/* use generate_series to generate a series of dates
   starting from the min date of datetime up to the 
   max datetime with one-month intervals, then do a 
   cross join with the distinct cust_id to map each cust_id 
   to each generated date.*/
WITH cust_dates AS
(
  SELECT EX.cust_id, to_char(dts, 'YYYY-mm') dts
  FROM generate_series 
        (
         (SELECT MIN(datetime)::timestamp FROM example), 
         (SELECT MAX(datetime)::timestamp + '2 month'::interval  FROM example),
         '1 month'::interval
        ) dts
  CROSS JOIN (SELECT DISTINCT cust_id FROM example) EX
),
/* do a left join with your table to find prices
   for each cust_id/ month, and aggregate for cust_id, month_date
   to find the sum of prices for each cust_id, month_date.
*/
monthly_price AS
(
  SELECT CD.cust_id,
       CD.dts AS month_date,
       COALESCE(SUM(price), 0) total_price
  FROM cust_dates CD LEFT JOIN example EX
  ON CD.cust_id = EX.cust_id AND 
     CD.dts = to_char(EX.datetime, 'YYYY-mm')
  GROUP BY CD.cust_id, CD.dts
)
/* Now, we have the sum of monthly prices for each cust_id,
   we can use the max window function with "ROWS BETWEEN 2 PRECEDING AND CURRENT ROW"
   to check if one of the (current month or the previous two months) has a sum of prices > 0.
*/
SELECT cust_id, month_date, total_price,
  CASE MAX(total_price) OVER 
      (PARTITION BY cust_id ORDER BY month_date 
       ROWS BETWEEN 2 PRECEDING AND CURRENT ROW)
    WHEN 0 THEN 'inactive' 
    ELSE 'active'
  END AS is_active
FROM monthly_price
ORDER BY cust_id, month_date

See demo