月底休息时间答案

【问题标题】：Break periods at the end of the month月底休息时间
【发布时间】：2021-04-05 05:19:49
【问题描述】：

SQL Server 2017

CREATE TABLE [TABLE_1] 
(
    PLAN_NR decimal(28,6) NULL,
    START_DATE datetime NULL,
);

INSERT INTO TABLE_1 (PLAN_NR, START_DATE) 
VALUES (1,'2020-05-01'), (2,'2020-08-01');

CREATE TABLE [TABLE_2] 
(
    PLAN_NR decimal(28,6) NULL,
    PERIOD_NR decimal(28,6) NOT NULL
);
   
INSERT INTO TABLE_2 (PLAN_NR, PERIOD_NR) 
VALUES (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8),
       (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8);

SQL-FIDDLE-LINK

TABLE_1 中有计划号和计划开始日期。

TABLE_2 包含每个计划编号的期间编号。

我想计算相应期间的开始日期：

每个周期正好是 7 天，除非该周期包含月末。那么这个时间段应该被划分为月底之前的范围，包括月底的最后一天和月底之后的范围。

选择：

SELECT 
    t1.PLAN_NR, t2.PERIOD_NR, 
    FORMAT(DATEADD (d ,((t2.PERIOD_NR-1)*7) , t1.START_DATE ),'yyyy-MM-dd') START_DATE
FROM
    TABLE_1 t1
JOIN
    TABLE_2 t2 ON t1.PLAN_NR = t2.PLAN_NR
ORDER BY 
    t1.PLAN_NR, t2.PERIOD_NR ASC

这将返回开始数据，但没有额外考虑相应的月末：

+---------+-----------+------------+
| PLAN_NR | PERIOD_NR | START_DATE |
+---------+-----------+------------+
|       1 |         1 | 2020-05-01 |
|       1 |         2 | 2020-05-08 |
|       1 |         3 | 2020-05-15 |
|       1 |         4 | 2020-05-22 |
|       1 |         5 | 2020-05-29 |
|       1 |         6 | 2020-06-05 |
|       1 |         7 | 2020-06-12 |
|       1 |         8 | 2020-06-19 |
|       2 |         1 | 2020-08-05 |
|       2 |         2 | 2020-08-12 |
|       2 |         3 | 2020-08-19 |
|       2 |         4 | 2020-08-26 |
|       2 |         5 | 2020-09-01 |
|       2 |         6 | 2020-09-02 |
|       2 |         7 | 2020-09-09 |
|       2 |         8 | 2020-09-16 |
+---------+-----------+------------+

我想要这样的输出：

+---------+-----------+----------------------+
| PLAN_NR | PERIOD_NR |      START_DATE      |
+---------+-----------+----------------------+
|       1 |         1 | 2020-05-01           |
|       1 |         2 | 2020-05-08           |
|       1 |         3 | 2020-05-15           |
|       1 |         4 | 2020-05-22           |
|       1 |         5 | 2020-05-29           |< --- period part before new month
|       1 |         6 | 2020-06-01           |< --- period part after new month
|       1 |         7 | 2020-06-05           |
|       1 |         8 | 2020-06-12           |
|       2 |         1 | 2020-08-05           |
|       2 |         2 | 2020-08-12           |
|       2 |         3 | 2020-08-19           |
|       2 |         4 | 2020-08-26           |< --- period part before new month
|       2 |         5 | 2020-09-01           |< --- period part after new month
|       2 |         6 | 2020-09-02           |
|       2 |         7 | 2020-09-09           |
|       2 |         8 | 2020-09-16           |
+---------+-----------+----------------------+

【问题讨论】：

您只想标记期间的结束/开始？
在您想要的结果示例中，带有句号的行。 8 和日期 2020-06-19 不存在。这是你的意图吗？如果是这样，它的逻辑是什么？请edit您的问题。
我这样做是因为我在所需的结果表中有错字，从 01.08 开始，而不是 opf 05.08
我想在 7 天期间内每次有月末时拆分 7 天期间
也许您可以尝试一个公用表表达式 (CTE)，它计算每 7 天期间的开始和结束日期，第二个 CTE 获取 Month( StartDate ) != Month( EndDate ) 所在的行并将它们分成两行（月末前后）和结合结果的最终查询。

标签： sql sql-server tsql datetime sql-server-2017

【解决方案1】：

使用窗口函数（LEAD / LAG）获取周期的开始和结束...

    SELECT t1.PLAN_NR
    , t2.PERIOD_NR
    , FORMAT(DATEADD (d ,((t2.PERIOD_NR-1)*7) , t1.START_DATE ),'yyyy-MM-dd') START_DATE
    , CASE
       WHEN 
        lead(
          FORMAT(DATEADD (d ,((t2.PERIOD_NR-1)*7) , t1.START_DATE ),'yyyy-MM-dd')
          ) over (partition by 
                    FORMAT(DATEADD (d ,((t2.PERIOD_NR-1)*7) , t1.START_DATE ),'yyyy-MM')
                  order by t2.period_nr)
         IS NULL THEN '< --- period part before new month'
       WHEN lag(
          FORMAT(DATEADD (d ,((t2.PERIOD_NR-1)*7) , t1.START_DATE ),'yyyy-MM-dd')
          ) over (partition by 
                    FORMAT(DATEADD (d ,((t2.PERIOD_NR-1)*7) , t1.START_DATE ),'yyyy-MM')
                  order by t2.period_nr)
         IS NULL THEN '< --- period part after new month'
       END as period_break
    from TABLE_1 t1
    join TABLE_2 t2
    on t1.PLAN_NR = t2.PLAN_NR
    order by t1.PLAN_NR, t2.PERIOD_NR asc

SQL Fiddle

PLAN_NR PERIOD_NR   START_DATE  period_break
1       1           2020-05-01  < --- period part after new month
1       2           2020-05-08  (null)
1       3           2020-05-15  (null)
1       4           2020-05-22  (null)
1       5           2020-05-29  < --- period part before new month
1       6           2020-06-05  < --- period part after new month
1       7           2020-06-12  (null)
1       8           2020-06-19  < --- period part before new month
2       1           2020-08-01  < --- period part after new month
2       2           2020-08-08  (null)
2       3           2020-08-15  (null)
2       4           2020-08-22  (null)
2       5           2020-08-29  < --- period part before new month
2       6           2020-09-05  < --- period part after new month
2       7           2020-09-12  (null)
2       8           2020-09-19  < --- period part before new month

【讨论】：

【解决方案2】：

SELECT 
    t1.PLAN_NR, t2.PERIOD_NR, 
    --row_number() over() but what if PERIOD_NR is not consecutive?
    t2.PERIOD_NR + SUM(num.n) OVER(PARTITION BY t2.PLAN_NR ORDER BY t2.PERIOD_NR, num.n) AS PERIOD_NR_x,
    FORMAT(CASE WHEN num.n = 1 THEN DATEADD(day, 1, EOMONTH(DATEADD (d ,((t2.PERIOD_NR-1)*7) , t1.START_DATE ))) ELSE DATEADD(d ,((t2.PERIOD_NR-1)*7) , t1.START_DATE ) END, 'yyyy-MM-dd') START_DATE
FROM
    TABLE_1 t1
JOIN
    TABLE_2 t2 ON t1.PLAN_NR = t2.PLAN_NR
CROSS APPLY
(
    SELECT 0 AS n
    UNION ALL
    --new row for month change
    SELECT 1 AS n
    WHERE DATEDIFF(month, DATEADD(d ,(t2.PERIOD_NR-1)*7 , t1.START_DATE), DATEADD(d ,t2.PERIOD_NR*7 , t1.START_DATE)) = 1
) as num    
ORDER BY 
    t1.PLAN_NR, t2.PERIOD_NR ASC

【讨论】：

那行得通。您在代码中的评论让我意识到可能存在没有连续和缺失的期间编号的情况。