Reputation: 25
I have a table as follows:
date table_name count_table
2018-07-19 A 50
2018-07-19 B 40
2018-07-18 A 25
2018-07-18 B 30
2018-07-17 A 10
2018-07-17 B 5
I would like to calculate the difference of the count between every date. The result table would be as:
date1 date2 table_name diff
2018-07-17 2018-07-18 A 15
2018-07-18 2018-07-19 A 25
2018-07-17 2018-07-18 B 25
2018-07-18 2018-07-19 B 10 `
Thank you for your help
Upvotes: 0
Views: 61
Reputation: 1173
If Multiple Dates and entities are there then you can use below code. i just written in MariaDb. You can use in any database by changing Temp table syntax.
CREATE TABLE test(dates date, table_name varchar(10), count_table INT);
INSERT INTO test(dates, table_name, count_table)
VALUES('2018-07-19','A', 50),
('2018-07-19', 'B', 40),
('2018-07-18', 'A', 25),
('2018-07-18', 'B', 30),
('2018-07-17', 'A', 10),
('2018-07-17', 'B', 5);
CREATE TEMPORARY TABLE test2(dates date, table_name varchar(10), count_table INT, Ranks INT);
INSERT INTO test2
select *, ROW_NUMBER() OVER(partition BY table_name order by dates) as Ranks
from test;
select t1.dates as date1, t2.dates as date2, t1.table_name, t2.count_table - t1.count_table
from test2 t1
INNER JOIN test2 t2 ON t1.table_name = t2.table_name and t1.ranks = t2.ranks - 1
ORDER BY table_name, t1.dates
Upvotes: 1
Reputation: 743
SELECT IT1.date1, ITM.date2, IT1.table_name, IT2.count_table - ITM.count_table AS diff
FROM input_table IT1
cross JOIN (
SELECT *
FROM input_table IT2
WHERE IT1.DATE < it2.DATE and IT1.table_name = IT2.table_name)
) ITM
Upvotes: 0
Reputation: 191
Use the same table twice in 1 query and give it an alias.
SELECT a.date AS date1, b.date AS date2, a.table_name AS table_name, b.count_table - a.count_table AS diff
FROM yourTable a, yourTable b
WHERE a.date1 < b.date2
GROUP BY date1, date2, table_name
Upvotes: 0