group by consecutive period - mysql

Question

Hi，I have a column as below
+--------+--------+
| day | amount|
+--------+---------
| 2 | 2 |
| 1 | 3 |
| 1 | 4 |
| 2 | 2 |
| 3 | 3 |
| 4 | 3 |
| 5 | 6 |
| 6 | 6 |
+--------+--------+
now I want something like this sum day 1- day3 as row one , sum day2-day4 as row 2, and so on.
+--------+--------+
| day | amount|
+--------+---------
| 1-3 | 14 |
| 2-4 | 10 |
| 3-5 | 12 |
| 4-6 | 15 |
+--------+--------+
Could you offer any one help ,thanks!

Answer 1

I would just use a correlated subquery:
select day, day + 2 as end_day,
(select sum(amount)
from t t2
where t2.day in (t.day, t.day + 1, t.day + 2)
) as amount
from (select distinct day from t) t;
This returns rows for all days, not limited to the last 4. If you really want that limit, then you can use:
select day, day + 2 as end_day,
(select sum(amount)
from t t2
where t2.day in (t.day, t.day + 1, t.day + 2)
) as amount
from (select distinct day
from t
order by day
offset 1 limit 99999999
) t
order by day;

Answer 2

Way 1:
Simply use UNION ALL:
SELECT '1 - 3' [Day], SUM(Amount)Amount FROM Your_Table WHERE Day BETWEEN 1 AND 3
UNION ALL
SELECT '2 - 4', SUM(Amount) FROM Your_Table WHERE Day BETWEEN 2 AND 4
UNION ALL
SELECT '3 - 5', SUM(Amount) FROM Your_Table WHERE Day BETWEEN 3 AND 5
UNION ALL
SELECT '4 - 6', SUM(Amount) FROM Your_Table WHERE Day BETWEEN 4 AND 6
Way 2:
You have to create a table with date range and JOIN the Table.
CREATE TABLE Tab1 (Day INT, Amount INT)
INSERT INTO Tab1 VALUES( 2 ,2 )
,(1, 3)
,(1, 4)
,(2, 2)
,(3, 3)
,(4, 3)
,(5, 6)
,(6, 6)
CREATE TABLE Tab2 (DateRange VARCHAR(10), StartDate INT, EndDate INT)
INSERT INTO Tab2 VALUES ('1 - 3',1,3)
,('2 - 4',2,4)
,('3 - 5',3,5)
,('4 - 6',4,6)
SELECT T2.DateRange,SUM(T1.Amount) Amount
FROM Tab1 T1
JOIN Tab2 T2 ON T1.Day BETWEEN T2.StartDate AND T2.EndDate
GROUP BY T2.DateRange
OutPut:
Day Amount
1 - 3 14
2 - 4 10
3 - 5 12
4 - 6 15

Answer 3

You can use integer division in order to calculate 'days buckets' and group by each bucket:
SELECT (day - 1) DIV 3 AS bucket, SUM(amount) AS total
FROM mytable
GROUP BY (day - 1) DIV 3;
Output:
bucket total
-------------
0 14
1 15
Demo here
To get the bucket string you can use:
SELECT concat(3 * ((day - 1) DIV 3 + 1) - 2, ' - ',
3 * ((day - 1) DIV 3 + 1)) AS bucket,
SUM(amount) AS total
FROM mytable
GROUP BY (day - 1) DIV 3
order by day;
Output:
bucket total
--------------
1 - 3 14
4 - 6 15
Note: The query works only for consecutive non-overlapping intervals.