Първо ще трябва да сумирате данните за месеца и след това да използвате функцията LAG, за да получите данни за предишни месеци, както следва:
SELECT
ORDER_MONTH,
LAG(UNIT_OF_PHONE_SALE, 1) OVER(
ORDER BY
ORDER_MONTH
) AS "M-1_Sale",
LAG(UNIT_OF_PHONE_SALE, 2) OVER(
ORDER BY
ORDER_MONTH
) AS "M-2_Sale",
LAG(UNIT_OF_PHONE_SALE, 3) OVER(
ORDER BY
ORDER_MONTH
) AS "M-3_Sale"
FROM
(
SELECT
TO_CHAR(ORDERDATE, 'YYYYMM') AS ORDER_MONTH,
SUM(UNIT_OF_PHONE_SALE) AS UNIT_OF_PHONE_SALE
FROM
DATAA
GROUP BY
TO_CHAR(ORDERDATE, 'YYYYMM')
)
ORDER BY
ORDER_MONTH DESC;
Изход:
ORDER_ M-1_Sale M-2_Sale M-3_Sale
------ ---------- ---------- ----------
201908 3789 789 666
201907 789 666 765
201906 666 765
201905 765
201904
Наздраве!!
- Актуализиране --
За изискването, споменато в коментарите, следната заявка ще работи за него.
CTE AS (
SELECT
TRUNC(ORDERDATE, 'MONTH') AS ORDER_MONTH,
SUM(UNIT_OF_PHONE_SALE) AS UNIT_OF_PHONE_SALE
FROM
DATAA
GROUP BY
TRUNC(ORDERDATE, 'MONTH')
)
SELECT
TO_CHAR(C.ORDER_MONTH,'YYYYMM') as ORDER_MONTH,
NVL(C1.UNIT_OF_PHONE_SALE, 0) AS "M-1_Sale",
NVL(C2.UNIT_OF_PHONE_SALE, 0) AS "M-2_Sale",
NVL(C3.UNIT_OF_PHONE_SALE, 0) AS "M-3_Sale"
FROM
CTE C
LEFT JOIN CTE C1 ON ( C1.ORDER_MONTH = ADD_MONTHS(C.ORDER_MONTH, - 1) )
LEFT JOIN CTE C2 ON ( C2.ORDER_MONTH = ADD_MONTHS(C.ORDER_MONTH, - 2) )
LEFT JOIN CTE C3 ON ( C3.ORDER_MONTH = ADD_MONTHS(C.ORDER_MONTH, - 3) )
ORDER BY
C.ORDER_MONTH DESC
Изход:
db<>демонстрация на fiddle от актуализиран отговор.
Наздраве!!