SQL 根据条件拆分岛屿

Question

我有一个 SQL table From 和 To 日期如下：

Row    From                    To 
--------------------------------------------------
1      2017-10-28 00:00:00     2017-10-30 00:00:00
2      2017-10-30 00:00:00     2017-10-31 00:00:00
3      2017-10-31 00:00:00     2017-10-31 07:30:00
4      2017-10-31 14:41:00     2017-10-31 15:14:00
5      2017-10-31 17:13:00     2017-11-01 00:00:00
6      2017-11-01 00:00:00     2017-11-01 23:45:00
7      2017-11-02 03:13:00     2017-11-02 07:56:00

我需要将连续的数据分组到孤岛中。数据不重叠。使用此查询可以轻松完成此操作：

;with Islands as
(
    SELECT
        min([From]) as [From]
        ,max([To]) as [To]
    FROM
    (
        select
            [From],
            [To],
            sum(startGroup) over (order by [From]) StartGroup
        from 
        (       
            SELECT 
                [From],
                [To],
                (case when [From] <= lag([To]) over (order by [From])
                                then 0
                                else 1
                        end) as StartGroup
                FROM dbo.DateTable
        ) IsNewIsland
    ) GroupedIsland
    group by StartGroup
)
select *
from Islands

然后给我这些结果：

From                    To                 Rows 
-----------------------------------------------------
2017-10-28 00:00:00     2017-10-31 07:30:00      1-3
2017-10-31 14:41:00     2017-10-31 15:14:00      4
2017-10-31 17:13:00     2017-11-01 23:45:00      5-6
2017-11-02 03:13:00     2017-11-02 07:56:00      7

我遇到的问题是，一旦岛屿获得足够的记录以达到一定的总持续时间，我就需要将查询修改为 cap/split 岛屿。这是一个 input/hardcoded 值。拆分包括整个记录，而不是在记录的 From-To 范围中间进行拆分。例如，我需要将岛屿拆分为 27 小时。这将给出以下结果：

From                    To                       Rows 
-----------------------------------------------------
2017-10-29 00:00:00     2017-10-30 00:00:00      1
2017-10-30 00:00:00     2017-10-31 07:30:00      2-3
2017-10-31 17:13:00     2017-11-01 23:45:00      5-6

第一个岛被拆分是因为第 1 行和第 2 行单独创建了 27 小时的时间段。第 4 行和第 7 行不足以创建一个岛，因此将其忽略。

我尝试通过内部 select 中的 lag 函数提取此信息来计算跨行的 "rolling duration"，但它不适用于跨越超过 2 行的岛屿因为它只会跟踪最后一行的持续时间，我无法"carry"向前计算。

SELECT 
  [From],
  [To],
  (case when [From] <= lag([To]) over (order by [From]
                            then (datediff(minute, [From], [To]) + lag(datediff(minute, [From], [To])) over (order by [From]))
                            else datediff(minute, [From], [To])
                    end) as RollingDuration,
  (case when [From] <= lag([To]) over (order by [From])
                      then 0
                      else 1
          end) as StartGroup
FROM dbo.DateTable

Answer 1

我能想到的 "least worst" 方法是 "quirky update"。 (Google这个，老实说不是我编的。)

http://www.sqlservercentral.com/articles/T-SQL/68467/
- 将数据复制到一个新的 table 中，其中包含一个或多个附加（空白）字段
- 使用 CLUSTERED PRIMARY KEY 确保以正确的顺序更新行
- 使用UPDATE和用户变量遍历行并存储计算结果

如果有间隙，或者运行总时长达到 27 小时，我可以使用它开始一个新组。然后照常进行。

-- New table to work through
----------------------------------------------------------------------
-- Addition [group_start] field (identifies groups, and useful data)
-- PRIMARY KEY CLUSTERED to enforce the order rows will be processed
----------------------------------------------------------------------

CREATE TABLE sample (
    id             INT,
    start          DATETIME,
    cease          DATETIME,
    group_start    DATETIME   DEFAULT(0),
    PRIMARY KEY CLUSTERED (group_start, start)   -- To force the order we will iterate the rows, and is useful in last step
);

INSERT INTO
    sample (
        id,
        start,
        cease
    )
VALUES
    (1,      '2017-10-28 00:00:00',     '2017-10-30 00:00:00'),
    (2,      '2017-10-30 00:00:00',     '2017-10-31 00:00:00'),
    (3,      '2017-10-31 00:00:00',     '2017-10-31 07:30:00'),
    (4,      '2017-10-31 14:41:00',     '2017-10-31 15:14:00'),
    (5,      '2017-10-31 17:13:00',     '2017-11-01 00:00:00'),
    (6,      '2017-11-01 00:00:00',     '2017-11-01 23:45:00'),
    (7,      '2017-11-02 03:13:00',     '2017-11-02 07:56:00')
;


-- Quirky Update
----------------------------------------------------------------------
-- Update [group_start] to the start of the current group
-- -> new group if gap since previous row
-- -> new group if previous row took group to 27 hours
-- -> else same group as previous row
----------------------------------------------------------------------

DECLARE @grp_start DATETIME = 0;

WITH
    lagged AS
(
    SELECT *, LAG(cease) OVER (ORDER BY group_start, start) AS lag_cease FROM sample
)
UPDATE
    lagged
SET
    @grp_start
        = group_start
            = CASE WHEN start <> lag_cease                     THEN start
                   WHEN start >= DATEADD(hour, 27, @grp_start) THEN start
                                                               ELSE @grp_start END
OPTION
    (MAXDOP 1)
;

-- Standard SQL to apply other logic
----------------------------------------------------------------------
-- MAX() OVER () to find end time of each group
-- WHERE to filter out any groups under 12 hours long
----------------------------------------------------------------------

SELECT
    *
FROM
(
    SELECT
        *,
        MAX(cease) OVER (PARTITION BY group_start)    AS group_cease
    FROM
        sample
)
   bounded_groups
WHERE
   group_cease >= DATEADD(hour, 12, group_start)
;

http://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=1bec5b3fe920c1affd58f23a11e280a0

SQL 根据条件拆分岛屿

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