根据 1 列的变化值对记录进行排名

Question

问：如何根据 1 列的变化值对记录进行排名？

我有以下数据（https://pastebin.com/vdTb1JRT）：

EmployeeID  Date        Onleave
ABH12345    2016-01-01  0
ABH12345    2016-01-02  0
ABH12345    2016-01-03  0
ABH12345    2016-01-04  0
ABH12345    2016-01-05  0
ABH12345    2016-01-06  0
ABH12345    2016-01-07  0
ABH12345    2016-01-08  0
ABH12345    2016-01-09  0
ABH12345    2016-01-10  1
ABH12345    2016-01-11  1
ABH12345    2016-01-12  1
ABH12345    2016-01-13  1
ABH12345    2016-01-14  0
ABH12345    2016-01-15  0
ABH12345    2016-01-16  0
ABH12345    2016-01-17  0

我想产生以下结果：

 EmployeeID DateValidFrom    DateValidTo     OnLeave
 ABH12345   2016-01-01       2016-01-09      0
 ABH12345   2016-01-10       2016-01-13      1
 ABH12345   2016-01-14       2016-01-17      0

所以我在考虑是否可以以某种方式创建一个排名列（如下所示），该列根据 Onleave 列中的值递增 - 由 EmployeeID 列分区。

EmployeeID  Date        Onleave    RankedCol
ABH12345    2016-01-01  0          1
ABH12345    2016-01-02  0          1
ABH12345    2016-01-03  0          1
ABH12345    2016-01-04  0          1
ABH12345    2016-01-05  0          1
ABH12345    2016-01-06  0          1
ABH12345    2016-01-07  0          1
ABH12345    2016-01-08  0          1
ABH12345    2016-01-09  0          1
ABH12345    2016-01-10  1          2
ABH12345    2016-01-11  1          2
ABH12345    2016-01-12  1          2
ABH12345    2016-01-13  1          2
ABH12345    2016-01-14  0          3
ABH12345    2016-01-15  0          3
ABH12345    2016-01-16  0          3
ABH12345    2016-01-17  0          3

那么我就可以做到以下几点：

SELECT
 [EmployeeID]    = [EmployeeID]
,[DateValidFrom] = MIN([Date])
,[DateValidTo]   = MAX([Date])
,[OnLeave]       = [OnLeave]
FROM table/view/cte/sub-query
GROUP BY 
 [EmployeeID]
,[OnLeave]
,[RankedCol]

非常欢迎其他解决方案..

以下是测试数据：

WITH CTE AS ( SELECT EmployeeID = 'ABH12345', [Date] = CAST(N'2016-01-01' AS Date), [Onleave] = 0
UNION SELECT 'ABH12345', CAST(N'2016-01-02' AS Date), 0
UNION SELECT 'ABH12345', CAST(N'2016-01-03' AS Date), 0
UNION SELECT 'ABH12345', CAST(N'2016-01-04' AS Date), 0
UNION SELECT 'ABH12345', CAST(N'2016-01-05' AS Date), 0
UNION SELECT 'ABH12345', CAST(N'2016-01-06' AS Date), 0
UNION SELECT 'ABH12345', CAST(N'2016-01-07' AS Date), 0
UNION SELECT 'ABH12345', CAST(N'2016-01-08' AS Date), 0
UNION SELECT 'ABH12345', CAST(N'2016-01-09' AS Date), 0
UNION SELECT 'ABH12345', CAST(N'2016-01-10' AS Date), 1
UNION SELECT 'ABH12345', CAST(N'2016-01-11' AS Date), 1
UNION SELECT 'ABH12345', CAST(N'2016-01-12' AS Date), 1
UNION SELECT 'ABH12345', CAST(N'2016-01-13' AS Date), 1
UNION SELECT 'ABH12345', CAST(N'2016-01-14' AS Date), 0
UNION SELECT 'ABH12345', CAST(N'2016-01-15' AS Date), 0
UNION SELECT 'ABH12345', CAST(N'2016-01-16' AS Date), 0
UNION SELECT 'ABH12345', CAST(N'2016-01-17' AS Date), 0
)

SELECT * FROM CTE

Answer 1

使用lag 的另一种方法。通过获取每个employeeid 的前一个Onleave 值并在找到不同值时重置它来分配组。

select employeeid,min(date) as date_from,max(date) as date_to,max(onleave) as onleave
from (select t.*,sum(case when prev_ol=onleave then 0 else 1 end) over(partition by employeeid order by date) as grp
      from (select c.*,lag(onleave,1,onleave) over(partition by employeeid order by date) as prev_ol
            from cte c
           ) t
      ) t
group by employeeid,grp 

Answer 2

这是获得所需输出的另一种更简单的方法 - 只访问表一次。

-- sample of data from your question
with t1(EmployeeID, Date1, Onleave) as(
  select 'ABH12345', cast('2016-01-01' as date),  0 union all
  select 'ABH12345', cast('2016-01-02' as date),  0 union all
  select 'ABH12345', cast('2016-01-03' as date),  0 union all
  select 'ABH12345', cast('2016-01-04' as date),  0 union all
  select 'ABH12345', cast('2016-01-05' as date),  0 union all
  select 'ABH12345', cast('2016-01-06' as date),  0 union all
  select 'ABH12345', cast('2016-01-07' as date),  0 union all
  select 'ABH12345', cast('2016-01-08' as date),  0 union all
  select 'ABH12345', cast('2016-01-09' as date),  0 union all
  select 'ABH12345', cast('2016-01-10' as date),  1 union all
  select 'ABH12345', cast('2016-01-11' as date),  1 union all
  select 'ABH12345', cast('2016-01-12' as date),  1 union all
  select 'ABH12345', cast('2016-01-13' as date),  1 union all
  select 'ABH12345', cast('2016-01-14' as date),  0 union all
  select 'ABH12345', cast('2016-01-15' as date),  0 union all
  select 'ABH12345', cast('2016-01-16' as date),  0 union all
  select 'ABH12345', cast('2016-01-17' as date),  0
)
-- actual query
select max(w.employeeid) as employeeid
     , min(w.date1)      as datevalidfrom
     , max(w.date1)      as datevalidto
     , max(w.onleave)    as onleave 
  from (
        select row_number() over(partition by employeeid order by date1) -
               row_number() over(partition by employeeid, onleave order by date1) as grp
             , employeeid
             , date1
             , onleave
          from t1 s
        ) w
group by w.grp
order by employeeid, datevalidfrom

结果：

employeeid datevalidfrom datevalidto onleave
---------- ------------- ----------- -----------
ABH12345   2016-01-01    2016-01-09  0
ABH12345   2016-01-10    2016-01-13  1
ABH12345   2016-01-14    2016-01-17  0

Answer 3

这是一个群岛问题的例子。在这种情况下，您可以使用日期算术。关键的观察结果是，从日期列中减去整数序列可以识别出相似值的孤岛。

作为查询，这看起来像：

SELECT EmployeeId, MIN([Date]) as DateValidFrom, MAX([Date]) as DateValidTo,
       OnLeave
FROM (SELECT t.*,
             ROW_NUMBER() OVER (PARTITION BY EmployeeId, OnLeave ORDER BY [Date]) as seqnum
      FROM t
     ) t
GROUP BY EmployeeID, DATEADD(day, - seqnum, [Date]), OnLeave;

您可以运行子查询，查看结果并进行算术运算，看看为什么会这样。

这里是example。