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
我想要从每组 PatientOID 中获取前 2 个 CollectedDT。
最佳答案
如果您至少使用 SQL-Server 2005,则可以使用 ROW_NUMBER
的 CTE
功能:
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
关于sql - 查询每组中的前 2 个,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11563574/