sql - 查找表中的最小和最大数据列

Question

我有一个表格，准确指定了每个员工在特定办公室的日期和时间。

EmployeeTable看起来像这样:

<表类=“s-表”> <标题> id 员工ID 日期P 时间P <正文> 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

我想为每个员工在办公室时准确选择最短和最长日期和时间:

<表类=“s-表”> <标题> id 员工ID 最小日期P 时间最小日期P 最大日期P TimeMaxDateP <正文> 1 11111 1397/01/02 01:30 1398/06/07 05:10 2 22222 1398/08/09 06:12 1399/02/01 07:15 3 11111 1399/07/02 08:51 1399/08/06 12:20 4 33333 1399/09/04 20:01 1400/02/05 14:10 5 22222 1400/04/05 16:25 1400/04/05 16:25

我的 SQL 代码是:

with tab1 as 
(
    select * 
    from EmployeeTable
), tab2 as 
(
    select 
        t1.*,
        case when lag(t1.EmployeeID) over(order by t1.id) is null then 1
            when lag(t1.EmployeeID) over(order by t1.id) = t1.EmployeeID 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.EmployeeID,
       min(t1.DateP) as min,
       TimeP,
       max(t1.DateP)as max,
       TimeP
  from tab3 t1
 group by t1.EmployeeID, t1.grp

但是上面的代码有错误。 每个人都可以帮助我吗？

Answer 1

这是一个间隙和岛屿问题。解决此问题的一种方法是使用行数差异法:

WITH cte AS (
    SELECT *, ROW_NUMBER() OVER (ORDER BY DateP, TimeP) rn1,
              ROW_NUMBER() OVER (PARTITION BY EmployeeID ORDER BY DateP, TimeP) rn2
    FROM EmployeeTable
),
cte2 AS (
    SELECT *, ROW_NUMBER() OVER (PARTITION BY EmployeeID, rn1 - rn2
                                 ORDER BY DateP, TimeP) rn_first,
              ROW_NUMBER() OVER (PARTITION BY EmployeeID, rn1 - rn2
                                 ORDER BY DateP DESC, TimeP DESC) rn_last
    FROM cte
)

SELECT
    EmployeeID,
    MAX(CASE WHEN rn_first = 1 THEN DateP END) AS MinDateP,
    MAX(CASE WHEN rn_first = 1 THEN TimeP END) AS TimeMinDateP,
    MAX(CASE WHEN rn_last = 1  THEN DateP END) AS MaxDateP,
    MAX(CASE WHEN rn_last = 1  THEN TimeP END ) AS TimeMaxDateP
FROM cte2
GROUP BY
    EmployeeID,
    rn1 - rn2
ORDER BY
    MIN(DateP),
    MIN(TimeP);

请注意，如果您使用单个日期时间列来表示日期和时间，则第二个 CTE 中的逻辑将完全没有必要。像您当前所做的那样将日期和时间分开通常没有好处。