How to GROUP BY with an exception?

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  C
1234567 2016-06-22  2016-06-22  C
1234567 2016-06-23  2016-06-23  D

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  2016-06-22  C
1234567 2016-06-23  2016-06-23  D

When I use min() and max() function with group by, it aggregates fine for all records but I have to aggregate only dates with the same ID day after day.

Any ideas?

Thanks in advance.

Answer 1

You can do it with conditional grouping by using CASE EXPRESSION in the GROUP BY clause :

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,
   CASE WHEN c.code in ('B','D') THEN b.a_day ELSE 1 END

Which will generate this as the GROUP BY clause :

c.code = 'B' -> a.user_id,c.code,b.a_day
c.code <> 'B' -> a.user_id,c.code,1