如何在不同时间获得相同ID的每个部分的最小日期？ - SQL 服务器

Posted 2023-03-16

【中文标题】如何在不同时间获得相同ID的每个部分的最小日期？ - SQL 服务器

【英文标题】：How to get minimum date by each section for same id different times? - SQL Server

【发布时间】：2017-06-01 12:15:53

【问题描述】：

我使用 SQL Server 2012。我有一些表，其中包含延迟总和的每日历史记录，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以及每个时期（部分）的结束日期） 细分市场相隔 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

暂时完成这项任务。 对不起我的英语。

【参考方案1】：

解决这个问题的一种方法是行号的差异：

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：

当我仔细观察时，您似乎想要第一个值，而不是最小值。 SQL Server 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);                   

【参考方案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 的记录分类到同一组中。此查询使用逻辑 dateadd(day,-row_number() over (partition by cid order by hdate),hdate) 来执行此操作。运行内部查询以查看组是如何分配的。

此后，使用窗口函数min、max 和first_value，您可以使用先前分配的组获得每个 cid 的最小 hdate、max hdate 和第一个值。

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