如何通过每个部分获得相同 ID 不同时间的最短日期？ - SQL 服务器

Question

我使用 SQL Server 2012。我有一些 table 的延迟总和每日历史记录，如下所示：

SET DATEFORMAT YMD
GO
    CREATE TABLE [dbo].[testsum](
        [CID] [int],
        [HDATE] [date],
        [DELAYSUM] [numeric](16, 2)
)
GO
INSERT [dbo].[testsum] ([CID], [HDATE], [DELAYSUM]) VALUES
        (223,'2016-10-16',15503.80)
        ,(223,'2016-10-17',15493.82)
        ,(223,'2016-10-18',15489.25)
        ,(223,'2016-10-19',15417.08)
        ,(427,'2016-10-01',10375.89)
        ,(427,'2016-10-02',10375.89)
        ,(427,'2016-10-03',10385.91)
        ,(427,'2016-10-16',8448.57)
        ,(427,'2016-10-17',8443.13)
        ,(427,'2016-10-18',8440.64)
        ,(427,'2016-10-19',8401.31)
        ,(427,'2016-10-20',8411.20)
        ,(427,'2016-10-21',8414.58)
        ,(427,'2016-10-22',8414.58)
        ,(427,'2016-10-23',8414.58)
        ,(427,'2016-10-24',8401.23)
        ,(427,'2016-10-25',8393.92)
        ,(427,'2016-10-26',8379.14)
        ,(427,'2016-10-27',8374.57)
        ,(427,'2016-10-28',8358.67)
        ,(427,'2016-10-29',8358.67)
        ,(427,'2016-10-30',8358.67)
        ,(427,'2016-10-31',8346.61)
        ,(541,'2016-10-05',900.44)
        ,(541,'2016-10-06',832.84)
        ,(541,'2016-10-11',637.54)
        ,(541,'2016-10-15',413.89)
        ,(541,'2016-10-16',413.89)
        ,(541,'2016-10-17',413.89)
        ,(541,'2016-10-18',1728.12)
        ,(541,'2016-10-22',265.27)
        ,(541,'2016-10-23',265.27)
        ,(541,'2016-10-24',265.27)
        ,(541,'2016-10-25',787.10)
        ,(541,'2016-10-26',1222.29)

10 月份 3 个 ID 的示例数据：

    CID         HDATE      DELAYSUM
    ----------- ---------- ---------------------------------------
    223       2016-10-16   15503.80
    223       2016-10-17   15493.82
    223       2016-10-18   15489.25
    223       2016-10-19   15417.08
    427       2016-10-01   10375.89
    427       2016-10-02   10375.89
    427       2016-10-03   10385.91
    427       2016-10-16   8448.57
    427       2016-10-17   8443.13
    427       2016-10-18   8440.64
    427       2016-10-19   8401.31
    427       2016-10-20   8411.20
    427       2016-10-21   8414.58
    427       2016-10-22   8414.58
    427       2016-10-23   8414.58
    427       2016-10-24   8401.23
    427       2016-10-25   8393.92
    427       2016-10-26   8379.14
    427       2016-10-27   8374.57
    427       2016-10-28   8358.67
    427       2016-10-29   8358.67
    427       2016-10-30   8358.67
    427       2016-10-31   8346.61
    541       2016-10-05   900.44
    541       2016-10-06   832.84
    541       2016-10-11   637.54
    541       2016-10-15   413.89
    541       2016-10-16   413.89
    541       2016-10-17   413.89
    541       2016-10-18   1728.12
    541       2016-10-22   265.27
    541       2016-10-23   265.27
    541       2016-10-24   265.27
    541       2016-10-25   787.10
    541       2016-10-26   1222.29

需要输出（ids (CID) 中每个日期部分的最小日期和每个期间（部分）的结束日期）分段间隔 1 天或更多天：

    CID         HDATE      DELAYSUM    END_DATE
    ----------- ---------- ---------------------------------------
    223       2016-10-16   15503.80    2016-10-19
    427       2016-10-01   10375.89    2016-10-03
    427       2016-10-16   8448.57     2016-10-31
    541       2016-10-05   900.44      2016-10-06
    541       2016-10-11   637.54      2016-10-11
    541       2016-10-15   413.89      2016-10-18
    541       2016-10-22   265.27      2016-10-26

暂时坚持这个任务。对不起我的英语。

Answer 1

解决这个问题的一种方法是行号的不同：

select cid, min(hdate), max(hdate), min(delaysum)
from (select t.*,
             row_number() over (order by hdate) as seqnum,
             row_number() over (partition by cid order by hdate) as seqnum_c  
      from testsum t
     ) t
group by cid, (seqnum - seqnum_c);

编辑：

仔细一看，您似乎想要第一个值，而不是最小值。 SQL 服务器不提供 first_value()（还）作为聚合函数。所以：

select cid, min(hdate), max(hdate), min(first_delaysum)
from (select t.*,
             first_value(delaysum) over (partition by cid, seqnum - seqnum_c order by hdate) as first_delaysum
      from (select t.*,
                   row_number() over (order by hdate) as seqnum,
                   row_number() over (partition by cid order by hdate) as seqnum_c   
            from testsum t
           ) t
     ) t
group by cid, (seqnum - seqnum_c);

Answer 2

这里的关键是如果连续行之间的日期差为1，则将每个cid的记录归为同一组。此查询使用逻辑dateadd(day,-row_number() over (partition by cid order by hdate),hdate)来完成。运行查看组分配方式的内部查询。

此后，使用 window 函数 min、max 和 first_value 您可以使用先前分配的组。

SELECT DISTINCT cid,
                min(hdate) over (partition BY cid, grp) AS hdate,
                first_value(delaysum) over (partition BY cid, grp ORDER BY hdate) AS delaysum,
                max(hdate) over (partition BY cid, grp) AS end_date
FROM (SELECT t.* ,
      dateadd(DAY,-row_number() over (partition BY cid ORDER BY hdate),hdate) AS grp
      FROM testsum t ) x
ORDER BY cid,hdate

Sample Demo

如何通过每个部分获得相同 ID 不同时间的最短日期？ - SQL 服务器

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