在sql server中按组选择每组的第一行

Question

我想像我在这里尝试的那样获得具有最小“差异”值的每个组的第一行。我怎样才能做到这一点？

此查询

SELECT 
ae.vpid as aeVpid,
ae.EreignisDat as aeEreignisDat, 
be.StichtagDat as beStichtagDat,
abs(datediff(day, ae.EreignisDat, be.StichtagDat)) as differenz
FROM AEGE as ae 
left outer join BEST as be
ON ae.VPID = be.VPID
WHERE ae.EreignisDat <> be.StichtagDat
Order By aevpid, differenz asc, beStichtagDat desc

明白我的意思

其中“beStichtagDat”是变量，并且取决于前两个属性“aeVpid”和“aeEreignisDat”

我在某些属性和子查询上尝试了一些group by，并在“differenz”上使用了min()，我还尝试了对组的row_number()，但我无法完成它

Answer 1

row_number()

+

partition by

可能是这里的方法，但这仍然不完整，因为问题显示了有关如何拆分组的任何内容：

SELECT  aeVpid, aeEreignisDat, beStichtagDat, differenz
FROM (
    select
        ae.vpid as aeVpid,
        ae.EreignisDat as aeEreignisDat, 
        be.StichtagDat as beStichtagDat,
        abs(datediff(day, ae.EreignisDat, be.StichtagDat)) as differenz,
        row_number() over (partition by ???? order by abs(datediff(day, ae.EreignisDat, be.StichtagDat))) rn

    from AEGE as ae 
    left join BEST as be on ae.VPID = be.VPID
    where ae.EreignisDat <> be.StichtagDat
) t
WHERE rn = 1
ORDER BY aevpid, differenz, beStichtagDat DESC