Question

我希望仅基于encounter,medicationname的群组获取所有数据列数据..

select encounter,medicationname,count(*) as freq,labdate,result 
from Medications where (labdate between @admitdate and DATEDIFF(dd,24,@admitdate)) 
group by encounter,medicationname having count(*)>2

我有像

这样的记录

encounter   medicationname  freq
8604261         ACC               3

现在根据这些数据，我想得到

这是我想要的输出

encounter   medicationname  labtime     result
8604261       ACC              2015-05-22    18
8604261       ACC              2015-07-23    23
8604261       ACC              2015-09-09    27

Answer 1

您可以使用PostDataAsync作为窗口函数，如下所示：

COUNT()

我会注意到我认为;With Counted as ( SELECT encounter,medicationname,labdate,result, COUNT(*) OVER (PARTITION BY encounter,medicationname) as cnt from Medications where (labdate between @admitdate and DATEDIFF(dd,24,@admitdate)) ) select encounter,medicationname,labdate,result from Counted where cnt > 2 ¹也可能是错误的，但由于我没有您的数据，输入和实际规格，我还是离开了就像现在一样。

¹ DATEDIFF会返回DATEDIFF，但您在与明显为int的列进行比较时会使用它。 date在这里可能是更可能需要的功能，但正如我所说，我没有完整的信息可以继续。

Answer 2

select encounter,medicationname,count(*) as freq,labdate,result 
from Medications 
where (labdate between @admitdate and DATEDIFF(dd,24,@admitdate)) 
group by encounter,medicationname having count(*) > 2

Answer 3

如果我理解你正确地问你需要的是这个

;WITH CTE AS 
(
select encounter,medicationname,count(*) as freq,labdate,result 
from Medications where (labdate between @admitdate and DATEDIFF(dd,24,@admitdate)) 
group by encounter,medicationname having count(*) > 2
)
select encounter,medicationname,labdate,result 
from Medications M
INNER JOIN CTE C
ON M.encounter = C.encounter
AND M.medicationname = C.medicationname
where (labdate between @admitdate and DATEDIFF(dd,24,@admitdate))

或更好地使用COUNT()OVER()

;WITH CTE AS
( 
    SELECT encounter,medicationname,COUNT(*) OVER(PARTITION BY encounter,medicationname)as freq,labdate,result 
    FROM Medications
    WHERE (labdate between @admitdate and DATEDIFF(dd,24,@admitdate)) 
)
SELECT * FROM CTE
WHERE freq > 2

select＆＃34; group by＆＃34;在特定列上，但显示其他列以及逐列

3 个答案: