SQL-分组依据，保留最早的一行

Question

I have following table:

name         team         date
-----------------------------------
John         A-team       02-5-2014
Jessica      A-team       08-6-2015
David        A-team       11-2-2013
Bill         B-team       12-5-2017
Nicole       B-team       18-1-2010
Brandom      B-team       22-9-2012

I am trying to create a query which does:

• one row per team, so we groupe on the team
• select that row which happened first, so we are aggregating on min(date)

The following query give the team and the date:

select   team, min(date)
from     my_table
group by team

But how can I also retrieve the name? I tried following query, but now I get all rows (which I understand, because the grouping does nothing, as all rows are unique now):

select   name, team, min(date)
from     my_table
group by team, name

Answer 1

Do a self join, where you use your first query as sub-query to find each team's first date:

select t1.*
from my_table t1
join
  (select team, min(date) min_date
   from my_table
   group by team) t2 on t1.team = t2.team and t1.date = t2.min_date

Will return both names if there are two names on a team's first date.

If you just want one row per team, even if that date has two name, you can do another GROUP BY , or simply NOT EXISTS :

select t1.*
from my_table t1
where not exists (select 1 from my_table t2
                  where t2.team = t1.team
                    and (t2.date < t1.date
                         or (t2.date = t1.date and t2.name < t1.name))

Answer 2

SELECT min(NAME)
    ,team
    ,min(DATE)
FROM my_table
GROUP BY team