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kyu6_conditional_count.sql
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-- Conditional Count;
--
-- https://www.codewars.com/kata/5816a3ecf54413a113000074
--
-- Given a payment table, which is a part of DVD Rental Sample Database, with the following schema
--
-- Column | Type | Modifiers
-- ---------+-----------------------------+----------
-- payment_id | integer | not null
-- customer_id | smallint | not null
-- staff_id | smallint | not null
-- rental_id | integer | not null
-- amount | numeric(5,2) | not null
-- payment_date | timestamp without time zone | not null
--
-- produce a result set for the report that shows a side-by-side comparison of the number and total amounts of payments
-- made in Mike's and Jon's stores broken down by months
--
-- The resulting data set should be ordered by month using natural order (Jan, Feb, Mar, etc.).
--
-- Note: You don't need to worry about the year component. Months are never repeated because the sample data set
-- contains payment information only for one year
--
-- The desired output for the report
-- month | total_count | total_amount | mike_count | mike_amount | jon_count | jon_amount
-- --+-------------+--------------+------------+-------------+-----------+-----------
-- 2 | | | | | |
-- 5 | | | | | |
--
-- month - number of the month (1 - January, 2 - February, etc.)
-- total_count - total number of payments
-- total_amount - total payment amount
-- mike_count - total number of payments accepted by Mike (staff_id = 1)
-- mike_amount - total amount of payments accepted by Mike (staff_id = 1)
-- jon_count - total number of payments accepted by Jon (staff_id = 2)
-- jon_amount - total amount of payments accepted by Jon (staff_id = 2)
SELECT EXTRACT(MONTH FROM payment_date) AS month,
COUNT(DISTINCT payment_id) as total_count,
SUM(amount) AS total_amount,
COUNT(CASE WHEN staff_id=1 THEN 1 ELSE NULL END) AS mike_count,
SUM(CASE WHEN staff_id=1 THEN amount ELSE NULL END) AS mike_amount,
COUNT(CASE WHEN staff_id=2 THEN 1 ELSE NULL END) AS jon_count,
SUM(CASE WHEN staff_id=2 THEN amount ELSE NULL END) AS jon_amount
FROM payment GROUP BY month;