從 SQL 中其他列的表基的每個部分中選擇最小和最大列值

Question

我有一個表格，顯示每個員工出現在一個辦公室時。 示例 SQL 表 EmpPTable 名稱是這樣的：

ID	員工ID	日期P	時間P
1	11111	1397/01/02	01:30
2	11111	1398/05/09	05:30
3	11111	1398/06/07	05:10
4	22222	1398/08/09	06:12
5	22222	1399/02/01	07:15
6	11111	1399/07/02	08:51
7	11111	1399/08/06	12:20
8	33333	1399/09/04	20:01
9	33333	1399/12/08	22:05
10	33333	1400/01/01	23:11
11	33333	1400/02/05	14:10
12	22222	1400/04/05	16:25

我想為每個在辦公室的員工准確選擇最小和最大日期：

ID	員工ID	最小日期	最大日期P
1	11111	1397/01/02	1398/06/07
2	22222	1398/08/09	1399/02/01
3	11111	1399/07/02	1399/08/06
4	33333	1399/09/04	1400/02/05
5	22222	1400/04/05	1400/04/05

我給出最大值的 SQL 代碼是：

       SELECT EmployeeID,MinDateP
        FROM (
          SELECT EmployeeID
          FROM EmpPTable
          GROUP BY EmployeeID
        ) As rstOuter
        OUTER APPLY (
          SELECT TOP 1 DateP As MinDateP
          FROM EmpPTable As rstInner
          WHERE rstInner.EmployeeID = rstOuter.EmployeeID
          ORDER BY DateP
        ) AS rstMinPoints

給出最小日期員工的 SQL 代碼是：

        SELECT EmployeeID,MaxDateP
        FROM (
          SELECT EmployeeID
          FROM EmpPTable
          GROUP BY EmployeeID
        ) As rstOuter
        OUTER APPLY (
          SELECT TOP 1 DateP As MaxDateP
          FROM EmpPTable As rstInner
          WHERE rstInner.EmployeeID = rstOuter.EmployeeID
          ORDER BY DateP desc
        ) AS rstMinPoints

但是上面的代碼為所有表中的每個員工提供了最小和最大日期列。

Answer 1

你應該group by列建立一個group by

mysql5.8

with tab1 as (
select 1 id, 1 user_id, '2021-11-01' dat from dual union all
select 2 id, 1 user_id, '2021-11-02' dat from dual union all
select 3 id, 1 user_id, '2021-11-03' dat from dual union all
select 4 id, 2 user_id, '2021-11-04' dat from dual union all
select 5 id, 2 user_id, '2021-11-05' dat from dual union all
select 6 id, 1 user_id, '2021-11-06' dat from dual union all
select 7 id, 1 user_id, '2021-11-07' dat from dual union all
select 8 id, 1 user_id, '2021-11-08' dat from dual union all
select 9 id, 3 user_id, '2021-11-09' dat from dual 
)
, tab2 as (
select t1.*,
       case when lag(t1.user_id) over(order by t1.id) is null then 1
            when lag(t1.user_id) over(order by t1.id) = t1.user_id then 0
            else 1
       end lg
  from tab1 t1
)
, tab3 as (
select t1.*,
       sum(t1.lg) over(order by t1.id) grp
  from tab2 t1
)
select t1.user_id,
       min(t1.dat),
       max(t1.dat)
  from tab3 t1
 group by t1.user_id, t1.grp

Answer 2

我無法復制您的代碼，因為缺少一些字段（serial1？）。 但這是我的解決方案。

CREATE TABLE #tempEmployee (
    ID INT,
    EmployeeID INT,
    EmpDate DATE,
    EmpTime TIME
);

INSERT INTO #tempEmployee VALUES (1, '11111', '2021-01-10', '12:30');
INSERT INTO #tempEmployee VALUES (1, '11111', '2021-12-10', '12:33');
INSERT INTO #tempEmployee VALUES (1, '11111', '2021-11-10', '12:30');
INSERT INTO #tempEmployee VALUES (2, '22222', '2021-01-11', '12:30');
INSERT INTO #tempEmployee VALUES (2, '22222', '2021-11-11', '12:30');

SELECT * FROM #tempEmployee;

WITH MaxDate AS (SELECT EmployeeID, MAX(EmpDate) AS MaxEmpDate FROM #tempEmployee GROUP BY EmployeeID),
     MinDate AS (SELECT EmployeeID, MIN(EmpDate) AS MinEmpDate FROM #tempEmployee GROUP BY EmployeeID)

     SELECT n.EmployeeID,
            n.MinEmpDate,
            x.MaxEmpDate
        FROM MinDate n 
        INNER JOIN MaxDate x ON n.EmployeeID = x.EmployeeID
        ORDER BY n.EmployeeID;

DROP TABLE #tempEmployee;

輸出....

EmployeeID  MinEmpDate  MaxEmpDate
11111       2021-01-10  2021-12-10
22222       2021-01-11  2021-11-11