请帮我生成以下查询。假设我有客户表和订单表。
客户表
CustID CustName
1 AA
2 BB
3 CC
4 DD
订单表
OrderID OrderDate CustID
100 01-JAN-2000 1
101 05-FEB-2000 1
102 10-MAR-2000 1
103 01-NOV-2000 2
104 05-APR-2001 2
105 07-MAR-2002 2
106 01-JUL-2003 1
107 01-SEP-2004 4
108 01-APR-2005 4
109 01-MAY-2006 3
110 05-MAY-2007 1
111 07-JUN-2007 1
112 06-JUL-2007 1
我想找出连续三个月订单的客户。 （允许使用SQL Server 2005和2008进行查询）。
所需的输出是：
CustName Year OrderDate
AA 2000 01-JAN-2000
AA 2000 05-FEB-2000
AA 2000 10-MAR-2000
AA 2007 05-MAY-2007
AA 2007 07-JUN-2007
AA 2007 06-JUL-2007
答案 0 :(得分：7)
编辑：摆脱或MAX() OVER (PARTITION BY ...)，因为这似乎会扼杀性能。
;WITH cte AS (
SELECT CustID ,
OrderDate,
DATEPART(YEAR, OrderDate)*12 + DATEPART(MONTH, OrderDate) AS YM
FROM Orders
),
cte1 AS (
SELECT CustID ,
OrderDate,
YM,
YM - DENSE_RANK() OVER (PARTITION BY CustID ORDER BY YM) AS G
FROM cte
),
cte2 As
(
SELECT CustID ,
MIN(OrderDate) AS Mn,
MAX(OrderDate) AS Mx
FROM cte1
GROUP BY CustID, G
HAVING MAX(YM)-MIN(YM) >=2
)
SELECT c.CustName, o.OrderDate, YEAR(o.OrderDate) AS YEAR
FROM Customers AS c INNER JOIN
Orders AS o ON c.CustID = o.CustID
INNER JOIN cte2 c2 ON c2.CustID = o.CustID and o.OrderDate between Mn and Mx
order by c.CustName, o.OrderDate
答案 1 :(得分：4)
这是我的版本。我真的把它作为一种纯粹的好奇心来展示，以展示另一种思考问题的方式。事实证明它比这更有用，因为它的表现甚至超过了马丁史密斯的酷“分组岛”解决方案。虽然，一旦他摆脱了一些过于昂贵的聚合窗口函数并做了真正的聚合，他的查询开始踢屁股。
解决方案1：运行3个月或更长时间，通过提前1个月检查并使用半联接来完成。
WITH Months AS (
SELECT DISTINCT
O.CustID,
Grp = DateDiff(Month, '20000101', O.OrderDate)
FROM
CustOrder O
), Anchors AS (
SELECT
M.CustID,
Ind = M.Grp + X.Offset
FROM
Months M
CROSS JOIN (
SELECT -1 UNION ALL SELECT 0 UNION ALL SELECT 1
) X (Offset)
GROUP BY
M.CustID,
M.Grp + X.Offset
HAVING
Count(*) = 3
)
SELECT
C.CustName,
[Year] = Year(OrderDate),
O.OrderDate
FROM
Cust C
INNER JOIN CustOrder O ON C.CustID = O.CustID
WHERE
EXISTS (
SELECT 1
FROM
Anchors A
WHERE
O.CustID = A.CustID
AND O.OrderDate >= DateAdd(Month, A.Ind, '19991201')
AND O.OrderDate < DateAdd(Month, A.Ind, '20000301')
)
ORDER BY
C.CustName,
OrderDate;
解决方案2：确切的3个月模式。如果是4个月或更长时间，则排除这些值。这是通过提前2个月和两个月后检查来完成的（主要是寻找N，Y，Y，Y，N模式）。
WITH Months AS (
SELECT DISTINCT
O.CustID,
Grp = DateDiff(Month, '20000101', O.OrderDate)
FROM
CustOrder O
), Anchors AS (
SELECT
M.CustID,
Ind = M.Grp + X.Offset
FROM
Months M
CROSS JOIN (
SELECT -2 UNION ALL SELECT -1 UNION ALL SELECT 0 UNION ALL SELECT 1 UNION ALL SELECT 2
) X (Offset)
GROUP BY
M.CustID,
M.Grp + X.Offset
HAVING
Count(*) = 3
AND Min(X.Offset) = -1
AND Max(X.Offset) = 1
)
SELECT
C.CustName,
[Year] = Year(OrderDate),
O.OrderDate
FROM
Cust C
INNER JOIN CustOrder O ON C.CustID = O.CustID
INNER JOIN Anchors A
ON O.CustID = A.CustID
AND O.OrderDate >= DateAdd(Month, A.Ind, '19991201')
AND O.OrderDate < DateAdd(Month, A.Ind, '20000301')
ORDER BY
C.CustName,
OrderDate;
如果有其他人想玩，这是我的表加载脚本：
IF Object_ID('CustOrder', 'U') IS NOT NULL DROP TABLE CustOrder
IF Object_ID('Cust', 'U') IS NOT NULL DROP TABLE Cust
GO
SET NOCOUNT ON
CREATE TABLE Cust (
CustID int identity(1,1) NOT NULL PRIMARY KEY CLUSTERED,
CustName varchar(100) UNIQUE
)
CREATE TABLE CustOrder (
OrderID int identity(100, 1) NOT NULL PRIMARY KEY CLUSTERED,
CustID int NOT NULL FOREIGN KEY REFERENCES Cust (CustID),
OrderDate smalldatetime NOT NULL
)
DECLARE @i int
SET @i = 1000
WHILE @i > 0 BEGIN
WITH N AS (
SELECT
Nm =
Char(Abs(Checksum(NewID())) % 26 + 65)
+ Char(Abs(Checksum(NewID())) % 26 + 97)
+ Char(Abs(Checksum(NewID())) % 26 + 97)
+ Char(Abs(Checksum(NewID())) % 26 + 97)
+ Char(Abs(Checksum(NewID())) % 26 + 97)
+ Char(Abs(Checksum(NewID())) % 26 + 97)
)
INSERT Cust
SELECT N.Nm
FROM N
WHERE NOT EXISTS (
SELECT 1
FROM Cust C
WHERE
N.Nm = C.CustName
)
SET @i = @i - @@RowCount
END
WHILE @i < 50000 BEGIN
INSERT CustOrder
SELECT TOP (50000 - @i)
Abs(Checksum(NewID())) % 1000 + 1,
DateAdd(Day, Abs(Checksum(NewID())) % 10000, '19900101')
FROM master.dbo.spt_values
SET @i = @i + @@RowCount
END
<强>性能
以下是3个月或更长时间查询的一些性能测试结果：
Query CPU Reads Duration
Martin 1 2297 299412 2348
Martin 2 625 285 809
Denis 3641 401 3855
Erik 1855 94727 2077
每个只有一次，但数字相当具有代表性。事实证明，你的查询表现并不是那么糟糕，毕竟丹尼斯。 Martin的查询击败了其他人，但起初他正在使用一些过于昂贵的窗口函数策略。
当然，正如我所指出的，当客户在同一天有两个订单时，Denis的查询并没有拉出正确的行，所以除非他修复了，否则他的查询是没有争用的。
此外，不同的指数可能会动摇。我不知道。
答案 2 :(得分：1)
你走了：
select distinct
CustName
,year(OrderDate) [Year]
,OrderDate
from
(
select
o2.OrderDate [prev]
,o1.OrderDate [curr]
,o3.OrderDate [next]
,c.CustName
from [order] o1
join [order] o2 on o1.CustId = o2.CustId and datediff(mm, o2.OrderDate, o1.OrderDate) = 1
join [order] o3 on o1.CustId = o3.CustId and o2.OrderId <> o3.OrderId and datediff(mm, o3.OrderDate, o1.OrderDate) = -1
join Customer c on c.CustId = o1.CustId
) t
unpivot
(
OrderDate for [DateName] in ([prev], [curr], [next])
)
unpvt
order by CustName, OrderDate
答案 3 :(得分：0)
这是我的看法。
select 100 as OrderID,convert(datetime,'01-JAN-2000') OrderDate, 1 as CustID into #tmp union
select 101,convert(datetime,'05-FEB-2000'), 1 union
select 102,convert(datetime,'10-MAR-2000'), 1 union
select 103,convert(datetime,'01-NOV-2000'), 2 union
select 104,convert(datetime,'05-APR-2001'), 2 union
select 105,convert(datetime,'07-MAR-2002'), 2 union
select 106,convert(datetime,'01-JUL-2003'), 1 union
select 107,convert(datetime,'01-SEP-2004'), 4 union
select 108,convert(datetime,'01-APR-2005'), 4 union
select 109,convert(datetime,'01-MAY-2006'), 3 union
select 110,convert(datetime,'05-MAY-2007'), 1 union
select 111,convert(datetime,'07-JUN-2007'), 1 union
select 112,convert(datetime,'06-JUL-2007'), 1
;with cte as
(
select
*
,convert(int,convert(char(6),orderdate,112)) - dense_rank() over(partition by custid order by orderdate) as g
from #tmp
),
cte2 as
(
select
CustID
,g
from cte a
group by CustID, g
having count(g)>=3
)
select
a.CustID
,Yr=Year(OrderDate)
,OrderDate
from cte2 a join cte b
on a.CustID=b.CustID and a.g=b.g