SQL min() max() function with exception - sql

Question

Here is my simplied code:
SELECT
a.user_id as User_ID,
min(b.a_day) as Date_from,
max(b.a_day) as Date_to,
c.code as ID
FROM a, b, c
WHERE
a_day > (day, -15, getdate())
GROUP BY
a.user_id,
c.code
Query gives the following output:
User ID date_from date_to id
1234567 2016-06-13 2016-06-13 B
1234567 2016-06-17 2016-06-17 A
12345672016-06-18 2016-06-18 A
1234567 2016-06-19 2016-06-19 A
1234567 2016-06-20 2016-06-20 A
1234567 2016-06-21 2016-06-21 B
I need something like this:
User ID date_from date_to id
1234567 2016-06-13 2016-06-13 B
1234567 2016-06-17 2016-06-20 A
1234567 2016-06-21 2016-06-21 B
When I use min() and max() function with group by, it aggregates fine for all records with ID=A but there should be exception for ID=B. I have to aggregate only dates with the same ID day after day.
Any ideas?
Thanks in advance.

Answer 1

You can combine these rows using the following strategy:
Determine where a new grouping begins.
Do a cumulative sum of the flag from (1) to identify each grouping.
Then do the aggregation.
This looks like:
select min(date_from) as date_from, max(date_to) as date_to, id
from (select t.*,
sum(isNewGroup) over (partition by id order by date_from) as grp
from (select t.*,
(case when lag(date_to) over (partition by id order by date_from) >= date_from
then 0 else 1
end) as isNewGroup
from t
) t
) t
group by id, grp;

Answer 2

it's my solution to get min/max continuous date.
try to run the SQL in your oracle.
is it helpful for you?
WITH TEST_DATA AS (
SELECT TO_DATE('20160613', 'YYYYMMDD') AS DATE_FROM, TO_DATE('20160613', 'YYYYMMDD') AS DATE_TO, 'B' AS ID FROM DUAL
UNION ALL
SELECT TO_DATE('20160617', 'YYYYMMDD') AS DATE_FROM, TO_DATE('20160617', 'YYYYMMDD') AS DATE_TO, 'A' AS ID FROM DUAL
UNION ALL
SELECT TO_DATE('20160618', 'YYYYMMDD') AS DATE_FROM, TO_DATE('20160618', 'YYYYMMDD') AS DATE_TO, 'A' AS ID FROM DUAL
UNION ALL
SELECT TO_DATE('20160619', 'YYYYMMDD') AS DATE_FROM, TO_DATE('20160619', 'YYYYMMDD') AS DATE_TO, 'A' AS ID FROM DUAL
UNION ALL
SELECT TO_DATE('20160620', 'YYYYMMDD') AS DATE_FROM, TO_DATE('20160620', 'YYYYMMDD') AS DATE_TO, 'A' AS ID FROM DUAL
UNION ALL
SELECT TO_DATE('20160621', 'YYYYMMDD') AS DATE_FROM, TO_DATE('20160621', 'YYYYMMDD') AS DATE_TO, 'B' AS ID FROM DUAL
)
SELECT
MIN(ID) AS ID,
MIN(DATE_FROM) AS DATE_FROM,
MAX(DATE_TO) AS DATE_TO
FROM (
SELECT
CONNECT_BY_ROOT(DATE_FROM) || CONNECT_BY_ROOT(ID) AS GROUP_KEY,
ROW_NUMBER() OVER(PARTITION BY ID, DATE_FROM, DATE_TO ORDER BY ID, LEVEL DESC) AS DISTINCT_FLG,
DATE_FROM,
DATE_TO,
ID
FROM
TEST_DATA
WHERE ID = CONNECT_BY_ROOT(ID)
CONNECT BY DATE_FROM = PRIOR DATE_TO + 1
ORDER BY DATE_FROM
)
WHERE
DISTINCT_FLG = 1
GROUP BY
GROUP_KEY

Answer 3

Here is mysql solution:
select grp, min(f) f, max(t) t, i
from
(
select x.*
,case when #lastu = i and datediff(f, #lastf)=1 then #gr:=#gr else #gr:=#gr+1 end grp
,#lastu:= i
,#lastf:= f
from
(
select '2016-06-13' f,'2016-06-13' t ,'B' i union all
select '2016-06-17','2016-06-17','A' union all
select '2016-06-18','2016-06-18','A' union all
select '2016-06-19','2016-06-19','A' union all
select '2016-06-20','2016-06-20','A' union all
select '2016-06-21','2016-06-21','B'
order by i, f, t
) x
, (select #gr:=0, #lastu:='', #lastf:='' ) b
) xx
group by grp, i