獲得數據

Question

在SQL Server組最小最大值，我有比的紀錄億盧比更像

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 

現在我想在那裏編組輸出，每個組都有最低最高值，因此輸出應該表像1的下方

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

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 2

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 

union all或union可以根據您的需要而定

Answer 3

現在也有一些變化，要求我要的是在同一個表中我 想是前兩名最大值

如果你使用SQL-Server，您可以使用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

如果你也想每empnos的MIN- /最大值，您可以使用OVER條款：

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 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 

結果：

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