在this algorithm 的改编版中，我们已经可以处理重叠：
declare @PersonA table(ClientID int, ID int, [From] date, Till date);
insert into @PersonA values (1,10,'20170101','20170430'),(1,12,'20170801','20180112');
declare @PersonB table(ClientID int, ID int, [From] date, Till date);
insert into @PersonB values (1,6,'20170301','20170630');
declare @PersonC table(ClientID int, ID int, [From] date, Till date);
insert into @PersonC values (1,12,'20170401','20170625');
declare @PersonD table(ClientID int, ID int, [From] date, Till date);
insert into @PersonD values (1,14,'20170501','20170525'),(1,14,'20170510','20171122');
with X(ClientID,EdgeDate)
as (select ClientID
,case
when toggle = 1
then Till
else [From]
end as EdgeDate
from
(
select ClientID,[From],Till from @PersonA
union all
select ClientID,[From],Till from @PersonB
union all
select ClientID,[From],Till from @PersonC
union all
select ClientID,[From],Till from @PersonD
) as concated
cross join
(
select-1 as toggle
union all
select 1 as toggle
) as toggler
),merged
as (select distinct
S.ClientID
,S.EdgeDate as [From]
,min(E.EdgeDate) as Till
from
X as S
inner join X as E
on S.ClientID = E.ClientID
and S.EdgeDate < E.EdgeDate
group by S.ClientID
,S.EdgeDate
),prds
as (select distinct
merged.ClientID
,merged.[From]
,merged.Till
,A.ID as PersonA
,B.ID as PersonB
,C.ID as PersonC
,D.ID as PersonD
from
merged
left join @PersonA as A
on merged.ClientID = A.ClientID
and A.[From] <= merged.[From]
and merged.Till <= A.Till
left join @PersonB as B
on merged.ClientID = B.ClientID
and B.[From] <= merged.[From]
and merged.Till <= B.Till
left join @PersonC as C
on merged.ClientID = C.ClientID
and C.[From] <= merged.[From]
and merged.Till <= C.Till
left join @PersonD as D
on merged.ClientID = D.ClientID
and D.[From] <= merged.[From]
and merged.Till <= D.Till
where not(A.ID is null
and B.ID is null
and C.ID is null
and D.ID is null
)
)
select ClientID
,[From]
,case
when Till = lead([From]
) over(order by Till)
then dateadd(d,-1,Till)
else Till
end as Till
,PersonA
,PersonB
,PersonC
,PersonD
from
prds
order by ClientID
,[From]
,Till;
仅包含问题中给出的两个 Person 表的输出：
+----------+------------+------------+---------+---------+
| ClientID | From | Till | PersonA | PersonB |
+----------+------------+------------+---------+---------+
| 1 | 2017-01-01 | 2017-02-28 | 10 | NULL |
| 1 | 2017-03-01 | 2017-04-29 | 10 | 6 |
| 1 | 2017-04-30 | 2017-06-30 | NULL | 6 |
| 1 | 2017-08-01 | 2018-01-12 | 12 | NULL |
+----------+------------+------------+---------+---------+
上面的脚本输出，有四个Person 表：
+----------+------------+------------+---------+---------+---------+---------+
| ClientID | From | Till | PersonA | PersonB | PersonC | PersonD |
+----------+------------+------------+---------+---------+---------+---------+
| 1 | 2017-01-01 | 2017-02-28 | 10 | NULL | NULL | NULL |
| 1 | 2017-03-01 | 2017-03-31 | 10 | 6 | NULL | NULL |
| 1 | 2017-04-01 | 2017-04-29 | 10 | 6 | 12 | NULL |
| 1 | 2017-04-30 | 2017-04-30 | NULL | 6 | 12 | NULL |
| 1 | 2017-05-01 | 2017-05-09 | NULL | 6 | 12 | 14 |
| 1 | 2017-05-10 | 2017-05-24 | NULL | 6 | 12 | 14 |
| 1 | 2017-05-25 | 2017-06-24 | NULL | 6 | 12 | 14 |
| 1 | 2017-06-25 | 2017-06-29 | NULL | 6 | NULL | 14 |
| 1 | 2017-06-30 | 2017-07-31 | NULL | NULL | NULL | 14 |
| 1 | 2017-08-01 | 2017-11-21 | 12 | NULL | NULL | 14 |
| 1 | 2017-11-22 | 2018-01-12 | 12 | NULL | NULL | NULL |
+----------+------------+------------+---------+---------+---------+---------+