本文介绍了查询以获取每组的前 2 名的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!
问题描述
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
这篇关于查询以获取每组的前 2 名的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持跟版网!
本站部分内容来源互联网,如果有图片或者内容侵犯您的权益请联系我们删除!