查询以获取每组的前 2 名

2021-09-10 00:00:00 sql tsql sql-server-2008 sql-server

我有这个sql代码:

SELECT   NoteOID = HCN.ObjectID,
         PatientOID = HCN.PatientID,
         PatientVisitOID = 0,
         CollectedDT = HCN.CollectedDT
FROM     HClinicalNote HCN WITH(NOLOCK)
where    HCN.enddt is NULL
         and HCN.visitid in (select distinct visitOID from @tblCensus)
order by HCN.PatientID,HCN.CollectedDT desc

给出这些结果:

 NoteOID    PatientOID     CollectedDT
 181382     890855         2011-09-14 21:31:00
 169115     890855         2011-09-12 18:38:00
 177466     890855         2011-09-09 19:49:00
 175150     890855         2011-09-07 19:34:00
 174057     890855         2011-09-06 19:25:00
 172429     890855         2011-09-04 09:00:00
 181387     13462666       2011-09-14 21:37:00
 182224     13462666       2011-09-14 13:24:00
 179269     13462666       2011-09-12 18:12:00

我想从每组 PatientOID 中获得前 2 个 CollectedDT.

I would like to have the top 2 CollectedDT from each group of PatientOID.

推荐答案

如果您至少使用 SQL-Server 2005，则可以使用 CTE 和 ROW_NUMBER 函数:

If you're using at least SQL-Server 2005, you ca use a CTE with ROW_NUMBER function:

WITH CTE AS(
   SELECT  NoteOID = HCN.ObjectID,
           PatientOID = HCN.PatientID,
           PatientVisitOID = 0,
           CollectedDT = HCN.CollectedDT,
           RN = ROW_NUMBER()OVER(PARTITION BY PatientOID ORDER BY CollectedDT ASC)
   FROM HClinicalNote HCN
   WHERE HCN.enddt is NULL
   AND HCN.visitid in (select distinct visitOID from @tblCensus)
)
SELECT * FROM CTE
WHERE RN <= 2

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