将连续的相似记录折叠成一条记录

Question

我有来自旧系统的人员的记录，我正试图将其转换为新系统。在旧系统中，一个人最终可能会在同一位置获得多条记录。他们也可以从一个位置移动到另一个位置，然后返回之前的位置。以下是一些示例数据：

PersonID  | LocationID | StartDate  | EndDate
1         | 1          | 1980-07-30 | 2007-07-16
1         | 1          | 2007-07-16 | 2008-01-30
1         | 2          | 2008-01-30 | 2009-03-02
1         | 2          | 2009-03-02 | 2009-11-06
1         | 3          | 2014-07-16 | 2015-01-16
1         | 1          | 2016-01-26 | 2999-12-31

我想折叠此数据，以便获得任何连续 LocationID 的日期范围。对于上面的数据，这是我所期望的：

PersonID  | LocationID | StartDate  | EndDate
1         | 1          | 1980-07-30 | 2008-01-30
1         | 2          | 2008-01-30 | 2009-11-06
1         | 3          | 2014-07-16 | 2015-01-16
1         | 1          | 2016-01-26 | 2999-12-31

我不确定如何执行此操作。我之前尝试加入上一条记录，但这仅在有两个连续位置时才有效，而不是 3 个或更多（可能存在未定义的连续记录数量）。

select
    a.PersonID,
    a.LocationID,
    a.StartDate,
    a.EndDate,
    case when a.LocationID = b.LocationID then a.PK_ID else b.PK_ID end as NewID
from employees a
left outer join employees b
on a.PersonID = b.PersonID
and a.PK_ID = b.PK_ID - 1

那么，如何编写查询来获得所需的结果？

注意：我们将“2999-12-31”视为我们的“NULL”日期字段

Answer 1

这是一个经典的峡谷和岛屿（编辑-更正了更大的跨度 2999）

Select [PersonID]
      ,[LocationID]
      ,[StartDate]  = min(D)
      ,[EndDate]    = max(D)
 From (
        Select *
              ,Grp = Row_Number() over (Order By D) - Row_Number() over (Partition By [PersonID],[LocationID] Order By D) 
         from YourTable A
         Cross Apply (
                        Select Top (DateDiff(DAY,A.[StartDate],A.[EndDate])+1) D=DateAdd(DAY,-1+Row_Number() Over (Order By (Select Null)),A.[StartDate])  
                        From  master..spt_values n1,master..spt_values n2
                     ) B
      ) G
 Group By [PersonID],[LocationID],Grp
 Order By [PersonID],min(D)

退货

PersonID    LocationID  StartDate   EndDate
1           1           1980-07-30  2008-01-30
1           2           2008-01-30  2009-11-06
1           3           2014-07-16  2015-01-16
1           1           2016-01-26  2999-12-31

使用您的原始查询

Select [PersonID]
      ,[LocationID]
      ,[StartDate]  = min(D)
      ,[EndDate]    = max(D)
 From (
        Select *
              ,Grp = Row_Number() over (Order By D) - Row_Number() over (Partition By [PersonID],[LocationID] Order By D) 
         From (
                -- Your Original Query
                select
                    a.PersonID,
                    a.LocationID,
                    a.StartDate,
                    a.EndDate,
                    case when a.LocationID = b.LocationID then a.PK_ID else b.PK_ID end as NewID
                from employees a
                left outer join employees b
                on a.PersonID = b.PersonID
                and a.PK_ID = b.PK_ID - 1
              ) A
         Cross Apply (
                        Select Top (DateDiff(DAY,A.[StartDate],A.[EndDate])+1) D=DateAdd(DAY,-1+Row_Number() Over (Order By (Select Null)),A.[StartDate])  
                        From  master..spt_values n1,master..spt_values n2
                     ) B
      ) G
 Group By [PersonID],[LocationID],Grp
 Order By [PersonID],min(D)

征求意见

让我们把它分解成它的组成部分。

1) CROSS APPLY 部分： 这会将单个记录扩展为 N 条记录。例如：

Declare @YourTable Table ([PersonID] int,[LocationID] int,[StartDate] date,[EndDate] date)
Insert Into @YourTable Values
 (1,1,'1980-07-01','1980-07-03' )
,(1,1,'1980-07-02','1980-07-04' )  -- Notice the Overlap
,(1,2,'2008-01-30','2008-02-05')

Select *
    from @YourTable A
    Cross Apply (
                Select Top (DateDiff(DAY,A.[StartDate],A.[EndDate])+1) D=DateAdd(DAY,-1+Row_Number() Over (Order By (Select Null)),A.[StartDate])  
                From  master..spt_values n1,master..spt_values n2
                ) B

上面的查询会生成

2) Grp 部分： 如果我提供一个简单的示例，可能会更容易：

Declare @YourTable Table ([PersonID] int,[LocationID] int,[StartDate] date,[EndDate] date)
Insert Into @YourTable Values
 (1,1,'1980-07-01','1980-07-03' )
,(1,1,'1980-07-02','1980-07-04' )  -- Notice the Overlap
,(1,2,'2008-01-30','2008-02-05')

Select *
      ,Grp = Row_Number() over (Order By D) - Row_Number() over (Partition By [PersonID],[LocationID] Order By D) 
      ,RN1 = Row_Number() over (Order By D)
      ,RN2 = Row_Number() over (Partition By [PersonID],[LocationID] Order By D) 
    from @YourTable A
    Cross Apply (
                Select Top (DateDiff(DAY,A.[StartDate],A.[EndDate])+1) D=DateAdd(DAY,-1+Row_Number() Over (Order By (Select Null)),A.[StartDate])  
                From  master..spt_values n1,master..spt_values n2
                ) B

以上查询生成：

RN1 和 RN2 是 GRP 的突破，只是为了说明机制。注意 RN1 减去 RN2 等于 GRP。一旦我们有了 GRP，就可以通过 group by 进行简单的聚合了

3) 齐心协力：

Declare @YourTable Table ([PersonID] int,[LocationID] int,[StartDate] date,[EndDate] date)
Insert Into @YourTable Values
 (1,1,'1980-07-01','1980-07-03' )
,(1,1,'1980-07-02','1980-07-04' )  -- Notice the Overlap
,(1,2,'2008-01-30','2008-02-05')

Select [PersonID]
      ,[LocationID]
      ,[StartDate]  = min(D)
      ,[EndDate]    = max(D)
 From (
        Select *
              ,Grp = Row_Number() over (Order By D) - Row_Number() over (Partition By [PersonID],[LocationID] Order By D) 
            from @YourTable A
            Cross Apply (
                        Select Top (DateDiff(DAY,A.[StartDate],A.[EndDate])+1) D=DateAdd(DAY,-1+Row_Number() Over (Order By (Select Null)),A.[StartDate])  
                        From  master..spt_values n1,master..spt_values n2
                        ) B
      ) G
 Group By [PersonID],[LocationID],Grp
 Order By [PersonID],min(D)

返回

Answer 2

对于您的样本数据，您可以使用行数差异的方法：

select personid, locationid, min(startdate), max(enddate)
from (select e.*,
             row_number() over (partition by personid order by startdate) as seqnum_p,
             row_number() over (partition by personid, locationid order by startdate) as seqnum_pl
      from employees  e
     ) e
group by (seqnum_p - seqnum_pl), personid, locationid;

这假定开始日期和结束日期是连续的。也就是说，同一地点的给定员工没有间隙。