SQL: Return only first occurrence

问题:

I seldomly use SQL and I cannot find anything similar in my archive so I'm asking this simple query question: I need a query which one returns personID and only the first seenTime

Records:

seenID | personID | seenTime
   108      3         13:34
   109      2         13:56
   110      3         14:22
   111      3         14:31
   112      4         15:04
   113      2         15:52

Wanted result:

personID | seenTime
   3         13:34
   2         13:56
   4         15:04

That's what I did & failed:

SELECT t.attendanceID, t.seenPersonID, t.seenTime
(SELECT ROW_NUMBER() OVER (PARTITION BY seenID ORDER BY seenID) AS RowNo,
seenID,
seenPersonID,
seenTime
FROM personAttendances) t
WHERE t.RowNo=1

P.S: Notice SQL CE 4

回答1:

If your seenTime increases as seenID increases:

select personID, min(seenTime) as seenTime
from personAttendances
group by personID

Update for another case:

If this is not the case, and you really want the seenTime that corresponds with the minimum seenID (assuming seenID is unique):

select a.personID, a.seenTime
from personAttendances as a
    join (
        -- Get the min seenID for each personID
        select personID, min(seenID) as seenID
        from personAttendances
        group by personID
    ) as b on a.personID = b.personID
where a.seenID = b.seenID

回答2:

You're making it way too difficult:

select personID, min(seenTime)
from personAttendances
group by personID

回答3:

You need to order by seen time not by seen id:

PARTITION BY seenID ORDER BY seenTime

回答4:

Add this to your SQL:

and where not exists
    (select 1 from personAttendances t2 
    where t.personID=t2.personID 
    and t2.seenID < t.seenID)