How to extract and pivot in sql - sql

Question

I have tables like following
I treid to sum score in pivoted style..
product date score
A 2020/8/1 1
B 2018/8/1 2
B 2018/9/1 1
C 2017/9/1 2
I'd like to transform them to the following pivotedone.
The index is YEAR(t.date) and columns = product
date A B C
2017 0 0 2
2018 0 3 0
2019 0 0 0
2020 1 0 0
Are there any effective way to achieve this?
Thanks

Answer 1

We can handle this by joining a calendar table containing all years of interest to your current table, aggregating by year, and then using conditional aggregation to find the sum of scores for each product.
WITH years AS (
SELECT 2017 AS year FROM dual UNION ALL
SELECT 2018 FROM dual UNION ALL
SELECT 2019 FROM dual UNION ALL
SELECT 2020 FROM dual
)
SELECT
y.year,
SUM(CASE WHEN t.product = 'A' THEN t.score ELSE 0 END) AS A,
SUM(CASE WHEN t.product = 'B' THEN t.score ELSE 0 END) AS B,
SUM(CASE WHEN t.product = 'C' THEN t.score ELSE 0 END) AS C
FROM years y
LEFT JOIN yourTable t
ON y.year = EXTRACT(YEAR FROM t."date")
GROUP BY
y.year
ORDER BY
y.year;
Demo

Answer 2

One option would be using PIVOT Clause after determining the year range, and joining outerly with your data source and setting the null scores as zeroes :
WITH years AS
(SELECT MIN(EXTRACT(year from "date")) AS min_year,
MAX(EXTRACT(year from "date")) AS max_year
FROM tab)
SELECT year, NVL(A,0) AS A, NVL(B,0) AS B, NVL(C,0) AS C
FROM (SELECT l.year, product, SUM(score) AS score
FROM tab t --> your original data source
RIGHT JOIN (SELECT level + min_year - 1 AS year
FROM years
CONNECT BY level BETWEEN 1 AND max_year - min_year + 1) l
ON l.year = EXTRACT(year from "date")
GROUP BY l.year, product)
PIVOT (SUM(score) FOR product IN('A' AS "A", 'B' AS "B", 'C' AS "C"))
ORDER BY year;
YEAR A B C
---- - - -
2017 0 0 2
2018 0 3 0
2019 0 0 0
2020 1 0 0
Demo