SQL: 获取同一列中行之间的日期差异

Question

我正在尝试创建报告，这是我的输入数据。

  Stage  Name         Date
    1     x     12/05/2019 10:00:03
    1     x     12/05/2019 10:05:01
    1     y     12/06/2019 12:00:07
    2     x     12/06/2019 13:12:03
    2     x     12/06/2019 13:23:00
    1     y     12/08/2019 16:00:07
    2     x     12/09/2019 09:17:59

这是我想要的输出。

  Stage  Name       DateFrom               DateTo            DateDiff
    1     x     12/05/2019 10:00:03    12/06/2019 12:00:07        1
    1     y     12/06/2019 12:00:07    12/06/2019 13:12:03        0
    2     x     12/06/2019 13:12:03    12/08/2019 16:00:07        2
    1     y     12/08/2019 16:00:07    12/09/2019 09:17:59        1

我不能在阶段和名称上使用 group by 子句，因为它会将我输入的第 3 行和第 6 行分组。我尝试将 table 加入自身，但没有得到想要的结果。这在 SQL 中甚至可能吗？任何想法都会有所帮助。我正在使用 Microsoft SQL 服务器。

Answer 1

这是间隙和孤岛问题的变体。您想要将相邻行的组组合在一起（即具有相同的阶段和名称）；但您想使用下一组的开始日期作为当前组的结束日期。

这是一种方法：

select 
    stage,
    name,
    min(date) date_from,
    lead(min(date)) over(order by min(date)) date_to,
    datediff(day, min(date), lead(min(date)) over(order by min(date))) date_diff 
from (
    select
        t.*,
        row_number() over(order by date) rn1,
        row_number() over(partition by stage, name order by date) rn2
    from mytable t
) t
group by stage, name, rn1 - rn2
order by date_from

Demo on DB Fiddle:

stage | name | date_from           | date_to             | datediff
----: | :--- | :------------------ | :------------------ | -------:
    1 | x    | 12/05/2019 10:00:03 | 12/06/2019 12:00:07 |        1
    1 | y    | 12/06/2019 12:00:07 | 12/06/2019 13:12:03 |        0
    2 | x    | 12/06/2019 13:12:03 | 12/08/2019 16:00:07 |        2
    1 | y    | 12/08/2019 16:00:07 | 12/09/2019 09:17:59 |        1
    2 | x    | 12/09/2019 09:17:59 | null                |     null

请注意，这不会完全您显示的结果：在结果集的末尾有一个额外的待处理记录，代表 "on-going"系列记录。如果需要，您可以通过嵌套查询将其过滤掉：

select *
from ( 
    select 
        stage,
        name,
        min(date) date_from,
        lead(min(date)) over(order by min(date)) date_to,
        datediff(day, min(date), lead(min(date)) over(order by min(date))) date_diff
    from (
        select
            t.*,
            row_number() over(order by date) rn1,
            row_number() over(partition by stage, name order by date) rn2
        from mytable t
    ) t
    group by stage, name, rn1 - rn2
) t 
where date_to is not null
order by date_from

Answer 2

这个是间隙和孤岛问题的变体，但它有一个非常简单的解决方案。

只需保留上一行具有不同阶段或名称的每一行。然后使用 lead() 获取下一个日期。这是基本思想：

select t.stage, t.name, t.date as datefrom
       lead(t.date) over (order by t.date) as dateto,
       datediff(day, t.date, lead(t.date) over (order by t.date)) as diff
from (select t.*,
             lag(date) over (partition by stage, name order by date) as prev_sn_date,
             lag(date) over (order by date) as prev_date
      from t
     ) t
where prev_sn_date <> prev_date or prev_sn_date is null;

如果真的要过滤掉最后一行，还需要多一步；我不确定这是否可取。

SQL: 获取同一列中行之间的日期差异

SQL: Get date difference between rows in the same column

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