आपको सबसे पहले महीने के डेटा का योग करना होगा और फिर पिछले महीनों के डेटा को निम्न के रूप में प्राप्त करने के लिए LAG फ़ंक्शन का उपयोग करना होगा:
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;
आउटपुट:
ORDER_ M-1_Sale M-2_Sale M-3_Sale
------ ---------- ---------- ----------
201908 3789 789 666
201907 789 666 765
201906 666 765
201905 765
201904
चीयर्स!!
-- अपडेट करें--
टिप्पणियों में उल्लिखित आवश्यकता के लिए, निम्नलिखित प्रश्न इसके लिए काम करेंगे।
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
आउटपुट:
db<>fiddle डेमो अद्यतन उत्तर का।
चीयर्स!!