DB2 分组和 Orderby

Question

I need the to order results by PUNCH_TS but retain the grouping of TYPE_CD .

Query:

select TYPE_CD, PUNCH_TS 
from PUNCH
group by PUNCH_TS, TYPE_CD 
order by TYPE_CD, PUNCH_TS asc

Result:

ADM 2020-01-13 12:00:00.0
ADM 2020-01-13 17:00:00.0
REG 2020-01-13 08:00:00.0
REG 2020-01-13 12:00:00.0

I changed the order by from the TYPE_CD to PUNCH_TS but then it loses the grouping of TYPE_CD :

REG 2020-01-13 08:00:00.0
ADM 2020-01-13 12:00:00.0
REG 2020-01-13 12:00:00.0
ADM 2020-01-13 17:00:00.0

What I need is:

REG 2020-01-13 08:00:00.0
REG 2020-01-13 12:00:00.0
ADM 2020-01-13 12:00:00.0
ADM 2020-01-13 17:00:00.0

Answer 1

This should do it:

order by TYPE_CD desc, PUNCH_TS

Rationale: if you want TYPE_CD 'REG' to appear before 'ADM' , then you need to a descending sort. The second sort criteria is ascending PUNCH_TS , so you get results ordered by ascending timestamp within groups having the same TYPE_CD .

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Edit

If you want the TYPE_CD that has the earlier timestamp first, then you can use window functions:

order by min(PUNCH_TS) over(partition by TYPE_CD), PUNCH_TS