[SQL]一组数据中Name列相同值的最大Je与最小je的差

declare @t table(name varchar(20),qy varchar(20),je int)
insert into @t 
select '产品一','北京',500
union all select '产品一','上海',300
union all select '产品二','北京',600
union all select '产品三','上海',1000
union all select '产品三','北京',8008
union all select '产品四','上海',400
--select * from @t a where not exists  --这是取表中的NAME相同的最大值
--(
--    select 1 from @t where name=a.name and je>a.je
--)
--第一个答案:
SELECT NAME,QY,JE-ISNULL((SELECT JE FROM @T WHERE NAME=A.NAME AND QY<>A.QY),0)
FROM @T A
WHERE JE=(SELECT MAX(JE) FROM @T WHERE NAME=A.NAME)
ORDER BY NAME
--第二个答案
Select 
      Name,
      qy,
      je-(Select 
                Case When Min(je)=A.je Then 0 Else Min(je) End 
          From 
                @t 
          Where Name=A.Name Group By Name) As je 
From 
      @t A 
Where Not Exists
     (Select 1 From @t Where Name=A.Name And je>A.je)
原文地址:https://www.cnblogs.com/beeone/p/3622289.html