在T-SQL中使用Over和Partition by時，Min（）上缺少日期

Question

我正在使用OVER和Partition by以獲取數據集的提示和最大日期。

|ResdetId | bookingdate | Amount | AmountExcl |
-----------------------------------------------
|120106   | 2018-02-04  |  75.00 |  70.7547   |
|120106   | 2018-02-05  |  75.00 |  70.7547   |
|120106   | 2018-02-06  |  90.00 |  84.9057   |
|120106   | 2018-02-08  |  75.00 |  70.7547   |
|120106   | 2018-02-09  |  75.00 |  70.7547   |

我正在使用此查詢

select distinct ResDetId, Amount, AmountExcl, 
    min(Bookingdate) OVER(Partition by ResDetId, Amount, AmountExcl) as Mindate,
    max(Bookingdate) OVER(Partition by ResDetId, Amount, AmountExcl) as MaxDate
 from @Cumulatedbookingdetails

我得到這個結果

|ResdetId | Amount | AmountExcl | MinDate    | MaxDate     |
------------------------------------------------------------
|120106   | 75.00  |  70.7547   | 2018-02-04 |  2018-02-09 |
|120106   | 90.00  |  84.9057   | 2018-02-06 |  2018-02-06 |

如我們所見，數據集中缺少日期2018-02-07記錄。 所以，我需要這樣的結果

|ResdetId | Amount | AmountExcl | MinDate    | MaxDate     |
------------------------------------------------------------
|120106   | 75.00  |  70.7547   | 2018-02-04 |  2018-02-05 |
|120106   | 75.00  |  70.7547   | 2018-02-08 |  2018-02-09 |
|120106   | 90.00  |  84.9057   | 2018-02-06 |  2018-02-06 |

Answer 1

解決此類“島嶼和缺口”問題的方法之一是使用遞歸CTE來建立島嶼。 我們使非遞歸部分（在union之上）找到標記每個島的起點的行，並且遞歸部分使每個島一次增長一個匹配項。

不幸的是，CTE的最終結果包含了用於構建孤島的所有中間行，因此您需要最終的GROUP來選擇最終的孤島：

declare @t table (ResdetId int, bookingdate date, Amount decimal(9,3), AmountExcl decimal (9,3))
insert into @t(ResdetId,bookingdate,Amount,AmountExcl) values
(120106,'20180204',75.00,70.7547),
(120106,'20180205',75.00,70.7547),
(120106,'20180206',90.00,84.9057),
(120106,'20180208',75.00,70.7547),
(120106,'20180209',75.00,70.7547)

;With Islands as (
    select ResdetId, Amount, AmountExcl,bookingdate as MinDate,bookingDate as MaxDate
    from @t t
    where not exists (select * from @t t2
        where t2.ResdetId = t.ResdetId
        and t2.Amount = t.Amount
        and t2.AmountExcl = t.AmountExcl
        and t2.bookingdate = DATEADD(day,-1,t.BookingDate))
    union all
    select i.ResdetId, i.Amount,i.AmountExcl,i.MinDate,t.bookingDate
    from Islands i
        inner join
        @t t
        on t.ResdetId = i.ResdetId
        and t.Amount = i.Amount
        and t.AmountExcl = i.AmountExcl
        and t.bookingdate = DATEADD(day,1,i.MaxDate)
)
select
    ResdetId, Amount, AmountExcl,MinDate,MAX(MaxDate) as MaxDate
from
    Islands
group by ResdetId, Amount, AmountExcl,MinDate

結果：

ResdetId    Amount    AmountExcl   MinDate    MaxDate
----------- --------- ------------ ---------- ----------
120106      75.000    70.755       2018-02-04 2018-02-05
120106      75.000    70.755       2018-02-08 2018-02-09
120106      90.000    84.906       2018-02-06 2018-02-06

Answer 2

由於預訂日期不在您的分區中，所以您沒有看到2018-02-07，因此

|ResdetId | Amount | AmountExcl 
--------------------------------
|120106   | 75.00  |  70.7547   
|120106   | 90.00  |  84.9057   

在您的分區中是唯一的。 就像鑰匙一樣。 您需要另一個屬性來區分相同的數據：

|ResdetId | Amount | AmountExcl 
--------------------------------
|120106   | 75.00  |  70.7547 

Answer 3

使用GROUP BY會容易得多。 OVER和DISTINCT是執行相同查詢的“更困難”的方法：

WITH VTE AS(
    SELECT ResdetId,
           CONVERT(date,bookingdate) AS bookingdate,
           Amount,
           AmountExcl
    FROM (VALUES (120106,'20180204',75.00,70.7547),
                 (120106,'20180205',75.00,70.7547),
                 (120106,'20180206',90.00,84.9057),
                 (120106,'20180208',75.00,70.7547),
                 (120106,'20180209',75.00,70.7547)) V(ResdetId,bookingdate,Amount,AmountExcl))
SELECT ResdetId,Amount,AmountExcl,
       MIN(bookingdate) AS MinBookingDate,
       MAX(bookingdate) AS MaxBookingDate
FROM VTE
GROUP BY ResdetId,Amount,AmountExcl;

正如我的薩米人所說，我讀錯了結果，這是一個空白和小島的問題：

WITH VTE AS(
    SELECT ResdetId,
           CONVERT(date,bookingdate) AS bookingdate,
           Amount,
           AmountExcl
    FROM (VALUES (120106,'20180204',75.00,70.7547),
                 (120106,'20180205',75.00,70.7547),
                 (120106,'20180206',90.00,84.9057),
                 (120106,'20180208',75.00,70.7547),
                 (120106,'20180209',75.00,70.7547)) V(ResdetId,bookingdate,Amount,AmountExcl)),
Grps AS(
    SELECT *,
           ROW_NUMBER() OVER (PARTITION BY ResdetId ORDER BY V.bookingdate) - 
           ROW_NUMBER() OVER (PARTITION BY ResdetId, Amount ORDER BY V.bookingdate) AS Grp
    FROM VTE V)
SELECT ResdetId,
       Amount,
       AmountExcl,
       MIN(bookingdate) AS MinBookingDate,
       MAX(bookingdate) AS MaxBookingDate
FROM Grps
GROUP BY ResdetId,
         Amount,
         AmountExcl,
         Grp
ORDER BY ResdetId,
         Amount,
         MinBookingDate;

Answer 4

試試這個，它使用行號差技術：

declare @tbl table(ResdetId int, bookingdate date, Amount float, AmountExcl float);
insert into @tbl values
(120106   , '2018-02-04'  ,  75.00 ,  70.7547   ),
(120106   , '2018-02-05'  ,  75.00 ,  70.7547   ),
(120106   , '2018-02-06'  ,  90.00 ,  84.9057   ),
(120106   , '2018-02-08'  ,  75.00 ,  70.7547   ),
(120106   , '2018-02-09'  ,  75.00 ,  70.7547   );

select MIN(bookingDate), MAX(bookingDate), Amount, AmountExcl
from (
    select *,
           ROW_NUMBER() over (order by bookingDate) -
           ROW_NUMBER() over (partition by amount, AmountExcl order by bookingDate) rn
    from @tbl
) a group by Amount, AmountExcl, rn