具有连续日期范围记录的数据集中具有最小和最大日期的日期范围

问题描述 投票：4回答：4

我有一个数据集，其中包含员工的身份，状态和日期范围。 下面给出的输入数据集是一名员工的详细信息。 记录中的日期范围是连续的（按照确切的顺序），使得第二行的开始日期将是第一行的结束日期的下一个日期。

如果员工连续休假不同月份，那么该表格会将日期范围内的信息存储为不同月份。 例如：在输入集中，员工已经从'16 -10-2016'到'31 -12-2016'休病假，并在'1-1-2017'加入。 所以此项目有3条记录，但日期是连续的。在输出中，我需要将其作为预期输出数据集中显示的一条记录。

INPUT

Id  Status   StartDate   EndDate

1   Active   1-9-2007    15-10-2016
1   Sick     16-10-2016  31-10-2016
1   Sick     1-11-2016   30-11-2016
1   Sick     1-12-2016   31-12-2016
1   Active   1-1-2017    4-2-2017  
1   Unpaid   5-2-2017    9-2-2017  
1   Active   10-2-2017   11-2-2017 
1   Unpaid   12-2-2017   28-2-2017 
1   Unpaid   1-3-2017    31-3-2017 
1   Unpaid   1-4-2017    30-4-2017 
1   Active   1-5-2017    13-10-2017
1   Sick     14-10-2017  11-11-2017
1   Active   12-11-2017  NULL

预期输出

Id   Status    StartDate    EndDate

1    Active    1-9-2007     15-10-2016
1    Sick      16-10-2016   31-12-2016
1    Active    1-1-2017     4-2-2017  
1    Unpaid    5-2-2017     9-2-2017  
1    Active    10-2-2017    11-2-2017 
1    Unpaid    12-2-2017    30-4-2017 
1    Active    1-5-2017     13-10-2017
1    Sick      14-10-2017   11-11-2017
1    Active    12-11-2017   NULL

我不能通过id，status取min（startdate）和max（EndDate）组，因为如果同一个员工已经采取了另一个病假，那么结束日期（示例中为'11 -11-2017'）将作为结束日期。

任何人都可以帮助我在SQL Server 2014中查询？

sql sql-server sql-server-2014
4个回答
1
投票

这是GROUPING和WINDOW的一个例子。

  • 首先，为每个状态设置一个重置点
  • 总和设置组
  • 然后获取每组的最大/最小日期。

;with x as
(
    select Id, Status, StartDate,  EndDate,
           iif (lag(Status) over (order by Id, StartDate) = Status, null, 1) rst
    from   emp
), y as
 (
    select Id, Status, StartDate, EndDate,
           sum(rst) over (order by Id, StartDate) grp
    from   x
 )

 select Id, 
        MIN(Status) as Status, 
        MIN(StartDate) StartDate, 
        MAX(EndDate) EndDate
 from   y
 group by Id, grp
 order by Id, grp

GO

Id | Status | StartDate           | EndDate            
-: | :----- | :------------------ | :------------------
 1 | Active | 01/09/2007 00:00:00 | 15/10/2016 00:00:00
 1 | Sick   | 16/10/2016 00:00:00 | 31/12/2016 00:00:00
 1 | Active | 01/01/2017 00:00:00 | 04/02/2017 00:00:00
 1 | Unpaid | 05/02/2017 00:00:00 | 09/02/2017 00:00:00
 1 | Active | 10/02/2017 00:00:00 | 11/02/2017 00:00:00
 1 | Unpaid | 12/02/2017 00:00:00 | 30/04/2017 00:00:00
 1 | Active | 01/05/2017 00:00:00 | 13/10/2017 00:00:00
 1 | Sick   | 14/10/2017 00:00:00 | 11/11/2017 00:00:00
 1 | Active | 12/11/2017 00:00:00 | null

dbfiddle here

3
投票

它突然袭击我，这基本上是一个gaps and islands问题 - 所以我完全改变了我的解决方案。 要使此解决方案起作用，日期不必是连续的。

首先，创建并填充样本表（请在将来的问题中保存此步骤）：

DECLARE @T AS TABLE
(
    Id int,
    Status varchar(10),
    StartDate date,
    EndDate date
);

SET DATEFORMAT DMY; -- This is needed because how you specified your dates.

INSERT INTO @T (Id, Status, StartDate, EndDate) VALUES
(1, 'Active', '1-9-2007', '15-10-2016'),
(1, 'Sick', '16-10-2016', '31-10-2016'),
(1, 'Sick', '1-11-2016', '30-11-2016'),
(1, 'Sick', '1-12-2016', '31-12-2016'),
(1, 'Active', '1-1-2017', '4-2-2017'),
(1, 'Unpaid', '5-2-2017', '9-2-2017'),
(1, 'Active', '10-2-2017', '11-2-2017'),
(1, 'Unpaid', '12-2-2017', '28-2-2017'),
(1, 'Unpaid', '1-3-2017', '31-3-2017'),
(1, 'Unpaid', '1-4-2017', '30-4-2017'),
(1, 'Active', '1-5-2017', '13-10-2017'),
(1, 'Sick', '14-10-2017', '11-11-2017'),
(1, 'Active', '12-11-2017',  NULL);

（新）公用表表达式：

;WITH CTE AS
(
SELECT  Id, 
        Status, 
        StartDate, 
        EndDate,
        ROW_NUMBER() OVER(PARTITION BY Id ORDER BY StartDate)
        - ROW_NUMBER() OVER(PARTITION BY Id, Status ORDER BY StartDate) As IslandId,
        ROW_NUMBER() OVER(PARTITION BY Id ORDER BY StartDate DESC)
        - ROW_NUMBER() OVER(PARTITION BY Id, Status ORDER BY StartDate DESC) As ReverseIslandId
FROM @T
)

（新）查询：

SELECT  DISTINCT Id,
        Status,
        MIN(StartDate) OVER(PARTITION BY IslandId, ReverseIslandId) As StartDate,
        NULLIF(MAX(ISNULL(EndDate, '9999-12-31')) OVER(PARTITION BY IslandId, ReverseIslandId), '9999-12-31') As EndDate

FROM CTE 
ORDER BY StartDate

（新）结果：

Id  Status  StartDate   EndDate
1   Active  01.09.2007  15.10.2016
1   Sick    16.10.2016  31.12.2016
1   Active  01.01.2017  04.02.2017
1   Unpaid  05.02.2017  09.02.2017
1   Active  10.02.2017  11.02.2017
1   Unpaid  12.02.2017  30.04.2017
1   Active  01.05.2017  13.10.2017
1   Sick    14.10.2017  11.11.2017
1   Active  12.11.2017  NULL

You can see a live demo on rextester.

请注意，SQL中日期的字符串表示应该符合ISO 8601 - 意思是yyyy-MM-dd或yyyyMMdd，因为它是明确的，并且将始终由SQL Server正确解释。

1
投票

这是一个不使用LAG的替代答案。

首先，我需要获取您的测试数据的副本：

DECLARE @table TABLE (Id INT, [Status] VARCHAR(50), StartDate DATE, EndDate DATE);
INSERT INTO @table SELECT 1, 'Active', '20070901', '20161015';
INSERT INTO @table SELECT 1, 'Sick', '20161016', '20161031';
INSERT INTO @table SELECT 1, 'Sick', '20161101', '20161130';
INSERT INTO @table SELECT 1, 'Sick', '20161201', '20161231';
INSERT INTO @table SELECT 1, 'Active', '20170101', '20170204';
INSERT INTO @table SELECT 1, 'Unpaid', '20170205', '20170209';
INSERT INTO @table SELECT 1, 'Active', '20170210', '20170211';
INSERT INTO @table SELECT 1, 'Unpaid', '20170212', '20170228';
INSERT INTO @table SELECT 1, 'Unpaid', '20170301', '20170331';
INSERT INTO @table SELECT 1, 'Unpaid', '20170401', '20170430';
INSERT INTO @table SELECT 1, 'Active', '20170501', '20171013';
INSERT INTO @table SELECT 1, 'Sick', '20171014', '20171111';
INSERT INTO @table SELECT 1, 'Active', '20171112', NULL;

然后查询是：

WITH add_order AS (
    SELECT
        *,
        ROW_NUMBER() OVER (ORDER BY StartDate) AS order_id
    FROM
        @table),
links AS (
    SELECT
        a1.Id,
        a1.[Status],
        a1.order_id,
        MIN(a1.order_id) AS start_order_id,
        MAX(ISNULL(a2.order_id, a1.order_id)) AS end_order_id,
        MIN(a1.StartDate) AS StartDate,
        MAX(ISNULL(a2.EndDate, a1.EndDate)) AS EndDate
    FROM
        add_order a1
        LEFT JOIN add_order a2 ON a2.Id = a1.Id AND a2.[Status] = a1.[Status] AND a2.order_id = a1.order_id + 1 AND a2.StartDate = DATEADD(DAY, 1, a1.EndDate)
    GROUP BY
        a1.Id,
        a1.[Status],
        a1.order_id),
merged AS (
    SELECT
        l1.Id,
        l1.[Status],
        l1.[StartDate],
        ISNULL(l2.EndDate, l1.EndDate) AS EndDate,
        ROW_NUMBER() OVER (PARTITION BY l1.Id, l1.[Status], ISNULL(l2.EndDate, l1.EndDate) ORDER BY l1.order_id) AS link_id
    FROM
        links l1
        LEFT JOIN links l2 ON l2.order_id = l1.end_order_id)
SELECT
    Id,
    [Status],
    StartDate,
    EndDate
FROM
    merged
WHERE
    link_id = 1
ORDER BY
    StartDate;

结果是：

Id  Status  StartDate   EndDate
1   Active  2007-09-01  2016-10-15
1   Sick    2016-10-16  2016-12-31
1   Active  2017-01-01  2017-02-04
1   Unpaid  2017-02-05  2017-02-09
1   Active  2017-02-10  2017-02-11
1   Unpaid  2017-02-12  2017-04-30
1   Active  2017-05-01  2017-10-13
1   Sick    2017-10-14  2017-11-11
1   Active  2017-11-12  NULL

它是如何工作的？首先，我添加一个序列号，以帮助将连续的行合并在一起。然后我确定可以合并在一起的行，添加一个数字来标识每个可以合并的集合中的第一行，最后从最终的CTE中挑出第一行。请注意，我还必须处理无法合并的行，因此需要处理LEFT JOINs和ISNULL语句。

只是为了感兴趣，这是最终CTE的输出看起来是什么，然后我过滤除了link_id为1的所有行：

Id  Status  StartDate   EndDate link_id
1   Active  2007-09-01  2016-10-15  1
1   Sick    2016-10-16  2016-12-31  1
1   Sick    2016-11-01  2016-12-31  2
1   Sick    2016-12-01  2016-12-31  3
1   Active  2017-01-01  2017-02-04  1
1   Unpaid  2017-02-05  2017-02-09  1
1   Active  2017-02-10  2017-02-11  1
1   Unpaid  2017-02-12  2017-04-30  1
1   Unpaid  2017-03-01  2017-04-30  2
1   Unpaid  2017-04-01  2017-04-30  3
1   Active  2017-05-01  2017-10-13  1
1   Sick    2017-10-14  2017-11-11  1
1   Active  2017-11-12  NULL        1
0
投票

您可以一起使用lag()和lead()函数来检查上一个和下一个状态

WITH CTE AS
(
    select  *, 
            COALESCE(LEAD(status) OVER(ORDER BY (select 1)), '0') Nstatus,
            COALESCE(LAG(status) OVER(ORDER BY (select 1)), '0') Pstatus
   from table
)

SELECT * FROM CTE
WHERE (status <> Nstatus AND status <> Pstatus) OR
      (status <> Pstatus)
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