SQL: Grouped average-if / case SELECT statement - mysql

Question

I have a database that looks like this SQL Fiddle: http://sqlfiddle.com/#!9/aa02e/1
CREATE TABLE Table1
(`Store` varchar(1), `Date` date, `Product` varchar(2), `Weekday` int, `Month` int, `Revenue` float)
;
INSERT INTO Table1
(`Store`, `Date`, `Product`, `Weekday`, `Month`, `Revenue`)
VALUES
('a', '20160101', 'aa', 5, 1, 1.5),
('a', '20160101', 'bb', 5, 1, 4),
('a', '20160101', 'cc', 5, 1, 3.5),
('a', '20160108', 'dd', 5, 1, 2.5),
('a', '20160108', 'ee', 5, 1, 5),
('b', '20160204', 'aa', 4, 2, 9.5),
('b', '20160204', 'bb', 4, 2, 4),
('b', '20160204', 'cc', 4, 2, 3),
('b', '20160211', 'dd', 4, 2, 1.5),
('b', '20160211', 'ee', 4, 2, 2.5)
;
SELECT * FROM table1;
+-------+------------+---------+---------+-------+---------+
| Store | Date | Product | Weekday | Month | Revenue |
+-------+------------+---------+---------+-------+---------+
| a | 2016-01-01 | aa | 5 | 1 | 1.5 |
| a | 2016-01-01 | bb | 5 | 1 | 4 |
| a | 2016-01-01 | cc | 5 | 1 | 3.5 |
| a | 2016-01-08 | dd | 5 | 1 | 2.5 |
| a | 2016-01-08 | ee | 5 | 1 | 5 |
| b | 2016-02-04 | aa | 4 | 2 | 9.5 |
| b | 2016-02-04 | bb | 4 | 2 | 4 |
| b | 2016-02-04 | cc | 4 | 2 | 3 |
| b | 2016-02-11 | dd | 4 | 2 | 1.5 |
| b | 2016-02-11 | ee | 4 | 2 | 2.5 |
+-------+------------+---------+---------+-------+---------+
It shows revenue data for stores incl. products, date and the respective day/month.
I want to select the following:
Store
Monthly revenue totals (i.e. what is the total revenue for store a in Jan?)
Weekday revenue averages (i.e. what is the avg revenue for store a on Thu?)
The first and second bullet are straightforward, but I'm having problems with the last one.
Currently, it takes the average over all products and all dates (assuming the weekday matches). What I need are the following steps:
Sum up all revenues for a store and a particular date (e.g. for store b: 9.5+4+3=16.5 for Feb 4th, and 1.5+2.5=4 for Feb 11th) if that date has the same weekday (here Thursday)
Take the average of the two values (e.g. avg(16.5,4)=10.25)
How can I accomplish that?
Thank you
Here is the query:
SELECT
Store,
SUM(CASE WHEN Month = 1 THEN Revenue ELSE NULL END) AS REVENUE_JAN,
SUM(CASE WHEN Month = 2 THEN Revenue ELSE NULL END) AS REVENUE_FEB,
AVG(CASE WHEN Weekday = 4 THEN Revenue ELSE NULL END) AS REVENUE_THU,
AVG(CASE WHEN Weekday = 5 THEN Revenue ELSE NULL END) AS REVENUE_FRI
FROM Table1
GROUP BY
Store
;

Answer 1

The weekday average is tricky. Your query is getting the average "order size" per weekday. But you want the total revenue.
One method is to first aggregate by weekday, but that is a bit of a mess. Instead, you can use this trick of calculating the average by dividing the total revenue by the number of days:
SELECT Store,
SUM(CASE WHEN Month = 1 THEN Revenue ELSE NULL END) AS REVENUE_JAN,
SUM(CASE WHEN Month = 2 THEN Revenue ELSE NULL END) AS REVENUE_FEB,
(SUM(CASE WHEN Weekday = 4 THEN Revenue END) /
COUNT(DISTINCT CASE WHEN Weekday = 4 THEN Date END)
) AS REVENUE_THU,
(SUM(CASE WHEN Weekday = 5 THEN Revenue END) /
COUNT(DISTINCT CASE WHEN Weekday = 5 THEN Date END)
) AS REVENUE_FRI
FROM Table1
GROUP BY Store;

Answer 2

SELECT
t1.store,
SUM(CASE WHEN Month = 1 THEN Revenue ELSE NULL END) AS REVENUE_JAN,
SUM(CASE WHEN Month = 2 THEN Revenue ELSE NULL END) AS REVENUE_FEB,
daily.REVENUE_THU,
daily.REVENUE_FRI
FROM Table1 t1
JOIN (
SELECT
Store,
weekday,
avg(CASE WHEN weekday = 4 THEN sum_rev END) as REVENUE_THU,
avg(CASE WHEN weekday = 5 THEN sum_rev END) as REVENUE_FRI
FROM (
SELECT
Store, date, weekday,
SUM(revenue) AS sum_rev
FROM Table1
GROUP BY
Store, date, weekday
) AS foo
GROUP BY Store, weekday
) AS daily ON daily.store = t1.store
GROUP BY
t1.store

Answer 3

How about this solution it return average for chosen day of chosen store
CREATE PROCEDURE sumForDayStore(IN vday INTEGER, IN vStore VARCHAR(50))
BEGIN
DECLARE totalDays INTEGER;
DECLARE totalRevenu INTEGER;
SET totalDays = (SELECT count(*) FROM Table1 WHERE WeekDay = vDay AND store = vStore);
SET totalRevenu = (SELECT sum(Revenue) FROM Table1 WHERE WeekDay = vDay AND store = vStore);
SELECT totalRevenu/totalDays;
END;
CALL sumForDayStore(5,'a');

Answer 4

How about this one:
SELECT mnth.Store, REVENUE_JAN, REVENUE_FEB, avg(rthu) REVENUE_THU, avg(rfri) REVENUE_FRI
FROM
(Select Store, sum(case when Month = 1 then Revenue else NULL END) REVENUE_JAN,
sum(case when Month = 2 then Revenue else NULL END) REVENUE_FEB
From Table1 group by Store) as mnth
join
(Select Store, sum(case when Weekday = 4 then Revenue end) rThu,
sum(case when Weekday = 5 then Revenue end) rFri from Table1 group by Store, Date) as dys
on mnth.Store = dys.Store
group by mnth.Store, REVENUE_JAN, REVENUE_FEB
I compared the performance of this with the query in the first answer and it shows better performance according to SQL server execution plan (1.6 times faster). Maybe this would be helpful on a larger data set.