10

使用 SQL Server 2008 R2，

鉴于一个结束日期紧邻下一个开始日期，我正在尝试将日期范围合并到最大日期范围中。

这些数据是关于不同的就业。一些员工可能已经结束了他们的工作并在以后重新加入。这些应该算作两种不同的工作（例如 ID 5）。有些人有不同类型的工作，彼此竞争（结束日期和开始日期并列），在这种情况下，它应该被视为一项工作（例如 ID 30）。

未结束的雇佣期的结束日期为空。

一些例子可能很有启发性：

declare @t as table  (employmentid int, startdate datetime, enddate datetime)

insert into @t values
(5, '2007-12-03', '2011-08-26'),
(5, '2013-05-02', null),
(30, '2006-10-02', '2011-01-16'),
(30, '2011-01-17', '2012-08-12'),
(30, '2012-08-13', null),
(66, '2007-09-24', null)

-- expected outcome
EmploymentId StartDate   EndDate
5            2007-12-03  2011-08-26
5            2013-05-02  NULL
30           2006-10-02  NULL
66           2007-09-24  NULL

我一直在尝试不同的“岛屿和间隙”技术，但无法破解这一技术。

4

4 回答 4

16

您在我使用日期“31211231”时看到的奇怪之处只是一个非常大的日期来处理您的“无结束日期”场景。我假设每个员工实际上不会有很多日期范围，所以我使用了一个简单的递归公用表表达式来组合这些范围。

为了让它运行得更快，起始锚查询只保留那些不会链接到先前范围（每个员工）的日期。其余的只是遍历日期范围并扩大范围。最后的 GROUP BY 仅保留每个起始 ANCHOR (employmentid, startdate) 组合建立的最大日期范围。

SQL小提琴

MS SQL Server 2008 架构设置：

create table Tbl (
  employmentid int,
  startdate datetime,
  enddate datetime);

insert Tbl values
(5, '2007-12-03', '2011-08-26'),
(5, '2013-05-02', null),
(30, '2006-10-02', '2011-01-16'),
(30, '2011-01-17', '2012-08-12'),
(30, '2012-08-13', null),
(66, '2007-09-24', null);

/*
-- expected outcome
EmploymentId StartDate   EndDate
5            2007-12-03  2011-08-26
5            2013-05-02  NULL
30           2006-10-02  NULL
66           2007-09-24  NULL
*/

查询 1：

;with cte as (
   select a.employmentid, a.startdate, a.enddate
     from Tbl a
left join Tbl b on a.employmentid=b.employmentid and a.startdate-1=b.enddate
    where b.employmentid is null
    union all
   select a.employmentid, a.startdate, b.enddate
     from cte a
     join Tbl b on a.employmentid=b.employmentid and b.startdate-1=a.enddate
)
   select employmentid,
          startdate,
          nullif(max(isnull(enddate,'32121231')),'32121231') enddate
     from cte
 group by employmentid, startdate
 order by employmentid

结果：

| EMPLOYMENTID |                        STARTDATE |                       ENDDATE |
-----------------------------------------------------------------------------------
|            5 |  December, 03 2007 00:00:00+0000 | August, 26 2011 00:00:00+0000 |
|            5 |       May, 02 2013 00:00:00+0000 |                        (null) |
|           30 |   October, 02 2006 00:00:00+0000 |                        (null) |
|           66 | September, 24 2007 00:00:00+0000 |                        (null) |
于 2013-04-03T09:32:42.487 回答
2 
SET NOCOUNT ON

DECLARE @T TABLE(ID INT,FromDate DATETIME, ToDate DATETIME)

INSERT INTO @T(ID,FromDate,ToDate)
SELECT 1,'20090801','20090803' UNION ALL
SELECT 2,'20090802','20090809' UNION ALL
SELECT 3,'20090805','20090806' UNION ALL
SELECT 4,'20090812','20090813' UNION ALL
SELECT 5,'20090811','20090812' UNION ALL
SELECT 6,'20090802','20090802'


SELECT ROW_NUMBER() OVER(ORDER BY s1.FromDate) AS ID,
       s1.FromDate, 
       MIN(t1.ToDate) AS ToDate 
FROM @T s1 
INNER JOIN @T t1 ON s1.FromDate <= t1.ToDate 
  AND NOT EXISTS(SELECT * FROM @T t2 
                 WHERE t1.ToDate >= t2.FromDate
                   AND t1.ToDate < t2.ToDate) 
WHERE NOT EXISTS(SELECT * FROM @T s2 
                 WHERE s1.FromDate > s2.FromDate
                   AND s1.FromDate <= s2.ToDate) 
GROUP BY s1.FromDate 
ORDER BY s1.FromDate
于 2018-03-16T04:32:41.267 回答
1

使用窗口函数而不是递归 CTE 的替代解决方案

SELECT 
    employmentid, 
    MIN(startdate) as startdate, 
    NULLIF(MAX(COALESCE(enddate,'9999-01-01')), '9999-01-01') as enddate
FROM (
    SELECT 
        employmentid, 
        startdate, 
        enddate,
        DATEADD(
            DAY, 
            -COALESCE(
                SUM(DATEDIFF(DAY, startdate, enddate)+1) OVER (PARTITION BY employmentid ORDER BY startdate ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING), 
                0
            ),
            startdate
    ) as grp
    FROM @t
) withGroup
GROUP BY employmentid, grp
ORDER BY employmentid, startdate

这通过计算grp对于所有连续行都相同的值来工作。这是通过以下方式实现的：

  1. 确定跨度占用的总天数（+1，因为包括日期）
SELECT *, DATEDIFF(DAY, startdate, enddate)+1 as daysSpanned FROM @t
  1. 每个工作的累计天数，按开始日期排序。这为我们提供了所有先前工作跨度的总天数
    • 我们与 0 合并以确保在我们的累积天数总和中没有 NULL
    • 我们在累积总和中不包括当前行，这是因为我们将使用该值startdate而不是enddate（我们不能使用它，enddate因为 NULL）
SELECT *, COALESCE(
    SUM(daysSpanned) OVER (
        PARTITION BY employmentid 
        ORDER BY startdate 
        ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING
    )
    ,0
)  as cumulativeDaysSpanned
FROM (
    SELECT *, DATEDIFF(DAY, startdate, enddate)+1 as daysSpanned FROM @t
) inner1
  1. 从 中减去累计天数startdate得到我们的grp. 这是解决方案的关键。
    • 如果开始日期的增长速度与所跨越的天数相同，那么这些天数是连续的，减去这两个天数将得到相同的值。
    • 如果 startdate 的增长速度快于所跨越的天数，则存在差距，我们将获得一个grp大于前一个值的新值。
    • 虽然grp是日期，但日期本身是没有意义的，我们只是将其用作分组值
SELECT *, DATEADD(DAY, -cumulativeDaysSpanned, startdate) as grp
FROM (
    SELECT *, COALESCE(
        SUM(daysSpanned) OVER (
            PARTITION BY employmentid 
            ORDER BY startdate 
            ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING
        )
        ,0
    )  as cumulativeDaysSpanned
    FROM (
        SELECT *, DATEDIFF(DAY, startdate, enddate)+1 as daysSpanned FROM @t
    ) inner1
) inner2

随着结果

+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| employmentid | startdate               | enddate                 | daysSpanned | cumulativeDaysSpanned | grp                     |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| 5            | 2007-12-03 00:00:00.000 | 2011-08-26 00:00:00.000 | 1363        | 0                     | 2007-12-03 00:00:00.000 |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| 5            | 2013-05-02 00:00:00.000 | NULL                    | NULL        | 1363                  | 2009-08-08 00:00:00.000 |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| 30           | 2006-10-02 00:00:00.000 | 2011-01-16 00:00:00.000 | 1568        | 0                     | 2006-10-02 00:00:00.000 |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| 30           | 2011-01-17 00:00:00.000 | 2012-08-12 00:00:00.000 | 574         | 1568                  | 2006-10-02 00:00:00.000 |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| 30           | 2012-08-13 00:00:00.000 | NULL                    | NULL        | 2142                  | 2006-10-02 00:00:00.000 |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| 66           | 2007-09-24 00:00:00.000 | NULL                    | NULL        | 0                     | 2007-09-24 00:00:00.000 |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
  1. 最后我们可以GROUP BY grp摆脱连续的日子。
    • 使用MINandMAX获取新的startdateandendate
    • 为了处理 NULL enddate，我们给它们一个很大的值来获取它们，然后再将MAX它们转换回NULL
SELECT 
    employmentid, 
    MIN(startdate) as startdate, 
    NULLIF(MAX(COALESCE(enddate,'9999-01-01')), '9999-01-01') as enddate
FROM (
    SELECT *, DATEADD(DAY, -cumulativeDaysSpanned, startdate) as grp
    FROM (
        SELECT *, COALESCE(
            SUM(daysSpanned) OVER (
                PARTITION BY employmentid 
                ORDER BY startdate 
                ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING
            )
            ,0
        )  as cumulativeDaysSpanned
        FROM (
            SELECT *, DATEDIFF(DAY, startdate, enddate)+1 as daysSpanned FROM @t
        ) inner1
    ) inner2
) inner3
GROUP BY employmentid, grp
ORDER BY employmentid, startdate

为了得到想要的结果

+--------------+-------------------------+-------------------------+
| employmentid | startdate               | enddate                 |
+--------------+-------------------------+-------------------------+
| 5            | 2007-12-03 00:00:00.000 | 2011-08-26 00:00:00.000 |
+--------------+-------------------------+-------------------------+
| 5            | 2013-05-02 00:00:00.000 | NULL                    |
+--------------+-------------------------+-------------------------+
| 30           | 2006-10-02 00:00:00.000 | NULL                    |
+--------------+-------------------------+-------------------------+
| 66           | 2007-09-24 00:00:00.000 | NULL                    |
+--------------+-------------------------+-------------------------+
  1. 我们可以结合内部查询来获取此答案开头的查询。哪个更短，但更难解释

所有这一切的限制要求

  • 就业的开始日期和结束日期没有重叠。这可能会在我们的grp.
  • 开始日期不为 NULL。然而，这可以通过用小日期值替换 NULL 开始日期来克服
  • 未来的开发者可以破译你执行的窗口黑魔法
于 2020-06-19T21:27:14.063 回答
0

用于组合所有重叠时段的修改脚本。
例如
01.01.2001-01.01.2010
05.05.2005-05.05.2015

将给出一个时期：
01.01.2001-05.05.2015

tbl.enddate 必须完成

;WITH cte
  AS(
SELECT
  a.employmentid
  ,a.startdate
  ,a.enddate
from tbl a
left join tbl c on a.employmentid=c.employmentid
    and a.startdate > c.startdate
    and a.startdate <= dateadd(day, 1, c.enddate)
WHERE c.employmentid IS NULL

UNION all

SELECT
  a.employmentid
  ,a.startdate
  ,a.enddate
from cte a
inner join tbl c on a.startdate=c.startdate
    and (c.startdate = dateadd(day, 1, a.enddate) or (c.enddate > a.enddate and c.startdate <= a.enddate))
)
select distinct employmentid,
          startdate,
          nullif(max(enddate),'31.12.2099') enddate
from cte
group by employmentid, startdate
于 2017-11-16T10:31:02.290 回答