SQL Server：優先順位が重複している日付の範囲を入力してください

Question

2017-04-01から2017-04-30までの範囲に、このテーブルのデータを入力する必要があります。私はこのような結果を取得する必要があります下の優先順位実際に

id  startValidity endValidity priority 
------------------------------------------- 
1004 2017-04-03 2017-04-30 1 
1005 2017-04-10 2017-04-22 2 
1010 2017-04-19 2017-04-23 3 
1006 2017-04-24 2017-04-28 2 
1008 2017-04-26 2017-04-28 3 

持つもの：

id  startValidity endValidity priority 
-------------------------------------------- 
1004 2017-04-03 2017-04-09 1 
1005 2017-04-10 2017-04-18 2 
1010 2017-04-19 2017-04-23 3 
1006 2017-04-24 2017-04-25 2 
1008 2017-04-26 2017-04-28 3 
1004 2017-04-29 2017-04-30 1 

Answer 1

は今エレガントなものか、より効率的な解決策を考えることはできません。 。 。

-- Sample Table 
declare @tbl table 
(
    id  int, 
    startValidity date, 
    endValidty date, 
    priority int 
) 

-- Sample Data 
insert into @tbl select 1004, '2017-04-03', '2017-04-30', 1 
insert into @tbl select 1005, '2017-04-10', '2017-04-22', 2 
insert into @tbl select 1010, '2017-04-19', '2017-04-23', 3 
insert into @tbl select 1006, '2017-04-24', '2017-04-28', 2 
insert into @tbl select 1008, '2017-04-26', '2017-04-28', 3 

-- Query 
; with 
date_range as -- find the min and max date for generating list of dates 
(
    select start_date = min(startValidity), end_date = max(endValidty) 
    from @tbl 
), 
dates as -- gen the list of dates using recursive CTE 
(
    select rn = 1, date = start_date 
    from date_range 

    union all 

    select rn = rn + 1, date = dateadd(day, 1, d.date) 
    from dates d 
    where d.date < (select end_date from date_range) 
), 
cte as -- for each date, get the ID based on priority 
(
    select *, grp = row_number() over(order by id) - rn 
    from dates d 
      outer apply 
      (
       select top 1 x.id, x.priority 
       from @tbl x 
       where x.startValidity <= d.date 
       and  x.endValidty >= d.date 
       order by x.priority desc 
      ) t 
) 
-- final result 
select id, startValidity = min(date), endValidty = max(date), priority 
from cte 
group by grp, id, priority 
order by startValidity 

Answer 2

私はカレンダーCTEやテーブルの目的を理解していません。 したがって、私はREcursive CTEまたはカレンダーを使用していません。

私は要件を完全に理解することはできません。

declare @tbl table 
(
    id  int, 
    startValidity date, 
    endValidty date, 
    priority int 
) 

-- Sample Data 
insert into @tbl select 1004, '2017-04-03', '2017-04-30', 1 
insert into @tbl select 1005, '2017-04-10', '2017-04-22', 2 
insert into @tbl select 1010, '2017-04-19', '2017-04-23', 3 
insert into @tbl select 1006, '2017-04-24', '2017-04-28', 2 
insert into @tbl select 1008, '2017-04-26', '2017-04-28', 3 
;With CTE as 
(
select * ,ROW_NUMBER()over(order by startValidity)rn 
from @tbl 
) 
,CTE1 as 
(
select c.id,c.startvalidity,isnull(dateadd(day,-1, c1.startvalidity) 
,c.endValidty) Endvalidity 
,c.[priority],c.rn 
from cte c 
left join cte c1 
on c.rn+1=c1.rn 
) 

select id,startvalidity,Endvalidity,priority from cte1 
union ALL 
select id,startvalidity,Endvalidity,priority from 
(
select top 1 id,ca.startvalidity,ca.Endvalidity,priority from cte1 
cross apply(
select top 1 
dateadd(day,1,endvalidity) startvalidity 
,dateadd(day,-1,dateadd(month, datediff(month,0,endvalidity)+1,0)) Endvalidity 
from cte1 
order by rn desc)CA 
order by priority 
)t4 
--order by startvalidity --if req