SQL Serverで計算日と計算日を取得する

Question

SQL Serverで以前の日付に関連するデータに不足している日付関連のデータを入力し、失われた日付の日付を取得する方法を教えてください。

表：

SELECT d.date, 
     o.orderid, 
     datediff(DAY, o.orderdate, d.date) AS missingdays, 
     o.cost 
FROM dateinfo d 
INNER JOIN 
    (SELECT o.orderid, 
      o.orderdate, 
      o.cost 
    FROM orders o) o ON o.orderdate <= d.date 
WHERE d.date BETWEEN '2016-06-01' AND '2016-06-09' 

が、クエリの上には、期待を返していません：私は

date  | orderid | missingdays | cost 
------------+---------+-------------+----- 
2016-06-01 | 10  |  0  | 100.00 
2016-06-02 | 11  |  0  | 200.00 
2016-06-02 | 14  |  0  | 700.00 
2016-06-03 | 11  |  1  | 200.00 
2016-06-03 | 14  |  1  | 700.00 
2016-06-04 | 11  |  2  | 200.00 
2016-06-04 | 14  |  2  | 700.00 
2016-06-05 | 12  |  0  | 300.00 
2016-06-06 | 12  |  1  | 300.00 
2016-06-07 | 12  |  2  | 300.00 
2016-06-08 | 12  |  3  | 300.00 
2016-06-09 | 13  |  0  | 400.00 
2016-06-09 | 15  |  0  | 700.00 

以下のようなデータは、私がこのようにしようとしたしたい上記のデータに基づいてdateinfo

CREATE TABLE [dbo].[dateinfo] 
(
    [date] [date] NULL 
) 
GO 

INSERT [dbo].[dateinfo] ([date]) 
VALUES (CAST(N'2016-06-01' AS Date)), 
     (CAST(N'2016-06-02' AS Date)), 
     (CAST(N'2016-06-03' AS Date)), 
     (CAST(N'2016-06-04' AS Date)), 
     (CAST(N'2016-06-05' AS Date)), 
     (CAST(N'2016-06-06' AS Date)), 
     (CAST(N'2016-06-07' AS Date)), 
     (CAST(N'2016-06-08' AS Date)), 
     (CAST(N'2016-06-09' AS Date)), 
     (CAST(N'2016-06-10' AS Date)), 
     (CAST(N'2016-06-11' AS Date)); 
go 

CREATE TABLE [dbo].[orders] 
(
    [orderid] [int] NULL, 
    [orderdate] [date] NULL, 
    [cost] [money] NULL 
) 
GO 

INSERT [dbo].[orders] ([orderid], [orderdate], [cost]) 
VALUES (10, CAST(N'2016-06-01' AS Date), 100.0000), 
     (11, CAST(N'2016-06-02' AS Date), 200.0000), 
     (12, CAST(N'2016-06-05' AS Date), 300.0000), 
     (13, CAST(N'2016-06-09' AS Date), 400.0000), 
     (14, CAST(N'2016-06-02' AS Date), 700.0000), 
     (15, CAST(N'2016-06-09' AS Date), 700.0000); 
GO 

結果。

SQL Serverでこの結果を達成するために、クエリを作成する方法を教えて下さい2012年

Answer 1

私はそれが簡単な方法で行うことができると思い、誰かのポストが良くなめらかになるまで、これを試してみてください。

declare @dateinfo table 
(
    [date] [date] NULL 
) 


INSERT @dateinfo ([date]) 
VALUES (CAST(N'2016-06-01' AS Date)), 
     (CAST(N'2016-06-02' AS Date)), 
     (CAST(N'2016-06-03' AS Date)), 
     (CAST(N'2016-06-04' AS Date)), 
     (CAST(N'2016-06-05' AS Date)), 
     (CAST(N'2016-06-06' AS Date)), 
     (CAST(N'2016-06-07' AS Date)), 
     (CAST(N'2016-06-08' AS Date)), 
     (CAST(N'2016-06-09' AS Date)), 
     (CAST(N'2016-06-10' AS Date)), 
     (CAST(N'2016-06-11' AS Date)); 
declare @orders table 
(
    [orderid] [int] NULL, 
    [orderdate] [date] NULL, 
    [cost] [money] NULL 
) 


INSERT @orders ([orderid], [orderdate], [cost]) 
VALUES (10, CAST(N'2016-06-01' AS Date), 100.0000), 
     (11, CAST(N'2016-06-02' AS Date), 200.0000), 
     (12, CAST(N'2016-06-05' AS Date), 300.0000), 
     (13, CAST(N'2016-06-09' AS Date), 400.0000), 
     (14, CAST(N'2016-06-02' AS Date), 700.0000), 
     (15, CAST(N'2016-06-09' AS Date), 700.0000); 



with c as 
( 
    select orderdate as cur, 
     lead(orderdate) over(order by orderdate) as nxt 
    from @orders 
), 

missing_ranges as 
(
select dateadd(day, 1, cur) as rangestart, 
     dateadd(day, -1, nxt) rangeend, 
     cur as prev_dt 
from c 
where datediff(day, cur, nxt) > 1 
) 

select d.date, 
     o.orderid, 
     datediff(day, o.orderdate, d.date) as missingdays, 
     o.cost 
from @dateinfo d 
    left join missing_ranges r 
     on d.date between r.rangestart and r.rangeend 
    left join @orders o 
     on d.date = o.orderdate or r.prev_dt = o.orderdate 
where d.date between '2016-06-01' and '2016-06-09'