Question

在sql server中，我有一个表格，其中包含的记录超过

empnos  date
a1  18-Jul-13
a1  18-Jul-13
a1  18-Jul-13
a1  18-Jul-13
a2  18-Jul-13
a2  18-Jul-13
a2  18-Jul-13
a3  18-Jul-13
a1  19-Jul-13
a2  19-Jul-13
a3  19-Jul-13
a1  20-Jul-13
a2  20-Jul-13
a3  20-Jul-13

现在我希望输出分组，每组都有最小值，所以输出应该像下面的1

EMPNO   DATE
A1  18-Jul-13
A1  20-Jul-13
A2  18-Jul-13
A2  20-Jul-13
A3  18-Jul-13
A3  20-Jul-13

Answer 1

现在我想要的一些变化在同一个表中想要是前两个最大值

如果您使用的是SQL-Server，可以使用ROW_NUMBER或DENSE_RANK：

WITH CTE AS(
   SELECT empnos,  
          date,
          rn = row_number() over (partition by empnos order by date desc)
   FROM dbo.TableName
)
SELECT * FROM CTE WHERE RN <= 2

Demo

如果您还需要每empnos的最小/最大值，则可以使用OVER子句：

WITH CTE AS(
   SELECT empnos,  
          date,
          min = Min(date) over (partition by empnos),
          max = Max(date) over (partition by empnos),
          rn = row_number() over (partition by empnos order by date desc)
   FROM dbo.TableName
)
SELECT * FROM CTE WHERE RN <= 2

Demo

Answer 2

Select empnos ,  Max(date) from TableName Group by empnos 
Union
Select empnos ,  Min(date) from TableName Group by empnos order by empnos

Sql Fiddle Demo

Answer 3

 select * from (
    Select empnos ,  Max(date) dat from TableName Group by empnos 
    Union all
    Select empnos ,  Min(date) dat from TableName Group by empnos 
 ) a 
    order by empnos, dat

可以根据您的要求使用

union all或union

Answer 4

DECLARE @t table (
   empnos   char(2)
 , the_date date
);

INSERT INTO @t (empnos, the_date)
  VALUES ('a1', '18-Jul-13')
       , ('a1', '18-Jul-13')
       , ('a1', '18-Jul-13')
       , ('a1', '18-Jul-13')
       , ('a2', '18-Jul-13')
       , ('a2', '18-Jul-13')
       , ('a2', '18-Jul-13')
       , ('a3', '18-Jul-13')
       , ('a1', '19-Jul-13')
       , ('a2', '19-Jul-13')
       , ('a3', '19-Jul-13')
       , ('a1', '20-Jul-13')
       , ('a2', '20-Jul-13')
       , ('a3', '20-Jul-13');

SELECT empnos
     , Max(the_date) As the_date
FROM   @t
GROUP
    BY empnos

UNION ALL

SELECT empnos
     , Min(the_date) As the_date
FROM   @t
GROUP
    BY empnos

ORDER
    BY empnos
     , the_date

结果：

empnos the_date
------ ----------
a1     2013-07-18
a1     2013-07-20
a2     2013-07-18
a2     2013-07-20
a3     2013-07-18
a3     2013-07-20

获取数据组中的最小最大值

4 个答案: