Sql server 如何返回按列A分组并在列B中具有不同值的记录
请原谅标题,但这是我在表Sql server 如何返回按列A分组并在列B中具有不同值的记录,sql-server,sql-server-2008,select,group-by,common-table-expression,Sql Server,Sql Server 2008,Select,Group By,Common Table Expression,请原谅标题,但这是我在表#RISK ACCT_NUM RISK 1234 5 007 10 007 1 37947 3 我希望输出为- ACCT_NUM RISK 007 10 007 1 这就是我尝试过的- ;WITH C AS( SELECT ROW_NUMBER() OVER (PARTITION BY ACCT_NUM ORDER BY ACCT_NUM, RISK) AS Rn
#RISK
ACCT_NUM RISK
1234 5
007 10
007 1
37947 3
我希望输出为-
ACCT_NUM RISK
007 10
007 1
这就是我尝试过的-
;WITH C AS(
SELECT ROW_NUMBER() OVER (PARTITION BY ACCT_NUM ORDER BY ACCT_NUM, RISK) AS Rn
,ACCT_NUM, RISK
FROM #RISK
WHERE NOT EXISTS(
SELECT ACCT_NUM,RISK FROM #RISK GROUP BY ACCT_NUM,RISK)
)
SELECT *
FROM C
但是我的查询没有返回任何结果。我做错了什么?这是你想要的吗
select acct_num, risk
from (select r.*,
min(risk) over (partition by acct_num) over minrisk,
max(risk) over (partition by acct_num) over maxrisk
from #risk r
) r
where minrisk <> maxrisk;
选择账户编号,风险
从(选择r.*),
min(风险)超过minrisk(按账户划分),
最大(风险)超过(按账户数划分)超过最大风险
从#风险r
)r
其中minrisk maxrisk;
我们也可以通过使用计数和行数函数来实现这一结果
declare @t table (AcctNum varchar(10),risk int)
insert into @t(AcctNum,risk)values
('1234',5),
('007',10),
('007',1),
('37947',3)
;with cte as (
select AcctNum,risk,COUNT(AcctNum)OVER(PARTITION BY AcctNum )RN from @t )
select AcctNum,risk from cte WHERE RN = 2
这有帮助吗
SELECT ACCT_NUM,RISK FROM #RISK
where ACCT_NUM=(
SELECT ACCT_NUM FROM #RISK
GROUP BY ACCT_NUM
HAVING COUNT(*)>1)