How to write the SQL for the following
ID PRODUCT LEAD_FLAG SALES_VOLUME
1 A Y 100
1 B N 200
1 C N 300
1 D N 400
2 A N 10
2 B Y 20
2 C N 30
2 D N 40
I need to calculate the incentive for each ID. The rule is for an ID:
if the the lead flag for a product is N, then check the sales_volume of the product whose lead_flag is Y (product A in case of ID 1, product B in case of ID 2). if the sales_volume of the product with lead_flag = Y is greater than a NUMBER (this NUMBER varies from product to product. for product A it is 20, for product B it is 25, C it is 30 and D it is 40) then incentive = (sales_volume of the product with lead flag N) * 100.
Hello, I presume the NUMBER you refer to will be held in a table column somewhere? I'm calling it threshold_num in the test data below (and rec_id instead of ID):
WITH test_data AS (
SELECT 1 REC_ID, 'A' PRODUCT, 'Y' LEAD_FLAG, 100 SALES_VOLUME FROM DUAL UNION ALL
SELECT 1, 'B','N', 200 FROM DUAL UNION ALL
SELECT 1, 'C','N', 300 FROM DUAL UNION ALL
SELECT 1, 'D','N', 400 FROM DUAL UNION ALL
SELECT 2, 'A','N', 10 FROM DUAL UNION ALL
SELECT 2, 'B','Y', 20 FROM DUAL UNION ALL
SELECT 2, 'C','N', 30 FROM DUAL UNION ALL
SELECT 2, 'D','N', 40 FROM DUAL),
test_ref_data AS (
SELECT 'A' PRODUCT, 20 threshold_num FROM DUAL UNION ALL
SELECT 'B', 25 FROM DUAL UNION ALL
SELECT 'C', 30 FROM DUAL UNION ALL
SELECT 'D', 40 FROM DUAL)
-- end test data
SELECT td1.REC_ID, td1.PRODUCT, CASE WHEN td2.PRODUCT IS NOT NULL THEN td1.sales_volume * 100 ELSE 0 END incentive
FROM test_data td1
LEFT JOIN (
SELECT td2.PRODUCT, SUM(sales_volume) sales_volume
FROM test_data td2
JOIN test_ref_data trd
ON (td2.PRODUCT = trd.PRODUCT)
WHERE td2.lead_flag = 'Y'
GROUP BY td2.PRODUCT, trd.threshold_num HAVING SUM(td2.sales_volume) > trd.threshold_num) td2
ON (td1.PRODUCT = td2.PRODUCT)
WHERE td1.lead_flag = 'N';
REC_ID PROD INCENTIVE
2 A 1000
2 D 0
1 D 0
1 B 0
2 C 0
1 C 0
6 rows selected.And two tips: it's always helps to put {noformat}{noformat} before and after your code for readability, and also to provide expected sample output.
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hi all,
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Tricky problem!
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Here's one solution:
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SELECT gpno, classnumber, age_min, age_max, amount
, COUNT (*) OVER ( PARTITION BY gpno
, classnumber
) AS gpno_classnumber_cnt
FROM fortest
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, a.classnumber
, MIN (b.classnumber)
OVER ( PARTITION BY a.gpno
, a.classnumber
) AS super_classnumber
FROM got_gpno_classnumber_cnt a
JOIN got_gpno_classnumber_cnt b ON a.gpno = b.gpno
AND a.age_min = b.age_min
AND a.age_max = b.age_max
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AND a.gpno_classnumber_cnt
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, got_rnk AS
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gpno, classnumber, super_classnumber
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SELECT DISTINCT
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AND c.classnumber = g.classnumber
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G123 1,3 30 35 2
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G123 2 36 40 5
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Check your solution below.
SQL> ed
Wrote file afiedt.buf
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