🚀 go-pugleaf

Now works!! only forcing PATIENTID to be equal:

SELECT
t.PATIENTID,
t.FECHA,
t.DIFF
FROM MYTABLE t
WHERE
t.DIFF = (SELECT MIN(q.DIFF) FROM 
MYTABLE q
WHERE t.PATIENTID = q.PATIENTID )




"aleplgr" <aleplgr@yahoo.es> wrote:
>
>Yes it's Interbase 7.5, sorry I posted here just a pseudocode, here's the code, now I get no error message but I'm not filtering the MIN DIFF, I'm getting all the patients, don't know why:
>
>SELECT
>t.PATIENTID,
>t.DATE,
>t.DIFF
>FROM MYTABLE t
>WHERE
>t.DIFF = (SELECT MIN(q.DIFF) FROM 
>MYTABLE q
>WHERE t.PATIENTID = q.PATIENTID AND t.DATE = q.DATE)
>
>
>
>
>
>"Karol Bieniaszewski" <liviuslivius@poczta.onet.pl> wrote:
>>
>>U�ytkownik "aleplgr" <aleplgr@yahoo.es> napisa� w wiadomo�ci 
>>news:462dca7d$1@newsgroups.borland.com...
>>>
>>> Hi, I have this problem: my  table is
>>> PATIENTID, DATE, DIFF
>>>
>>> for each PATIENTID can be many DATE and many DIFF, DIFF is an integer and 
>>> I need for each PATIENTID the MIN DIFF and the DATE of the same line where 
>>> MIN(DIFF) is.
>>>
>>> I did this: but it's not working
>>>
>>> SELECT PATIENTID,DATE,DIFF FROM T MYTABLE
>>> WHERE DIFF = (SELECT MIN(DIFF) FROM Q MYTABLE
>>>      WHERE (T.PATIENTID = Q.PATIENTID
>>>           AND   T.DATE = Q.DATE )
>>>
>>
>>Is this Interbase?
>>In Interbase alias is after table name.
>>And what errror do you get?
>>
>>Karol Bieniaszewski 
>>
>>
>