我需要在不使用循环和相关子查询的情况下使用SQL生成下表中的“required”列。这在SQL 2008中是否可行？
Date Customer Value Required Rule
20100101 1 12 12
20100101 2 0 If no value assign 0
20100101 3 32 32
20100101 4 42 42
20100101 5 15 15
20100102 1 12 Take last known value
20100102 2 0 Take last known value
20100102 3 39 39
20100102 4 42 Take last known value
20100102 5 16 16
20100103 1 13 13
20100103 2 24 24
20100103 3 39 Take last known value
20100103 4 42 Take last known value
20100103 5 21 21
20100104 1 14 14
20100104 2 24 Take last known value
20100104 3 39 Take last known value
20100104 4 65 65
20100104 5 23 23
基本上我正在为该客户填充具有最后知道值的空“值”单元格。请记住，最后一行可能没有有效值，因此您必须从之前的行中选择有效值。
答案 0 :(得分：5)
法伊兹，
以下查询怎么样，根据我的理解，它会做你想要的。评论解释了每一步。看看联机丛书上的CTE。甚至可以更改此示例以使用SQL 2008的新MERGE命令。
/* Test Data & Table */
DECLARE @Customers TABLE
(Dates datetime,
Customer integer,
Value integer)
INSERT INTO @Customers
VALUES ('20100101', 1, 12),
('20100101', 2, NULL),
('20100101', 3, 32),
('20100101', 4, 42),
('20100101', 5, 15),
('20100102', 1, NULL),
('20100102', 2, NULL),
('20100102', 3, 39),
('20100102', 4, NULL),
('20100102', 5, 16),
('20100103', 1, 13),
('20100103', 2, 24),
('20100103', 3, NULL),
('20100103', 4, NULL),
('20100103', 5, 21),
('20100104', 1, 14),
('20100104', 2, NULL),
('20100104', 3, NULL),
('20100104', 4, 65),
('20100104', 5, 23) ;
/* CustCTE - This gives us a RowNum to allow us to build the recursive CTE CleanCust */
WITH CustCTE
AS (SELECT Customer,
Value,
Dates,
ROW_NUMBER() OVER (PARTITION BY Customer ORDER BY Dates) RowNum
FROM @Customers),
/* CleanCust - A recursive CTE. This runs down the list of values for each customer, checking the Value column, if it is null it gets the previous non NULL value.*/
CleanCust
AS (SELECT Customer,
ISNULL(Value, 0) Value, /* Ensure we start with no NULL values for each customer */
Dates,
RowNum
FROM CustCte cur
WHERE RowNum = 1
UNION ALL
SELECT Curr.Customer,
ISNULL(Curr.Value, prev.Value) Value,
Curr.Dates,
Curr.RowNum
FROM CustCte curr
INNER JOIN CleanCust prev ON curr.Customer = prev.Customer
AND curr.RowNum = prev.RowNum + 1)
/* Update the base table using the result set from the recursive CTE */
UPDATE trg
SET Value = src.Value
FROM @Customers trg
INNER JOIN CleanCust src ON trg.Customer = src.Customer
AND trg.Dates = src.Dates
/* Display the results */
SELECT * FROM @Customers
答案 1 :(得分：2)
我需要制作专栏 “必需”在下表中 使用SQL而不使用循环和 相关的子查询。这是 在SQL 2008中可能吗？
不可能。点。不可能在任何基于SQL的服务器上，包括oracle。
这里的主要问题是您排除循环和相关子查询，并且在查询时检索值的任何方式最终将使用另一个查询来查找有效值（实际上每个字段一个）。这就是SQL的工作原理。是的，您可以在自定义标量函数中隐藏它们，但它们仍然包含逻辑子查询。
答案 2 :(得分：2)
这是＆＃34; Last non-null puzzle，＆＃34;这是几个优雅解决方案之一：
如果你的＆＃34;稀疏＆＃34; table为SparseTable，其中包含Date，Customer，Value列：
with C as
(select *,
max(case when Value is not null then [Date] end)
over (partition by Customer order by [Date] rows unbounded preceding) as grp
from SparseTable
)
insert into FullTable
select *,
max(Value) over (partition by Customer, grp order by [Date] rows unbounded preceding) as Required
from C
Value无法向前填充的地方仍为NULL，因此您可以
update FullTable set Required = 0 where Required is null
答案 3 :(得分：1)
我不确定以下是否考虑到你的约束，但它完成了工作。
测试数据
DECLARE @Customers TABLE (Date DATETIME, Customer INTEGER, Value INTEGER)
INSERT INTO @Customers VALUES ('20100101', 1, 12 )
INSERT INTO @Customers VALUES ('20100101', 2, NULL)
INSERT INTO @Customers VALUES ('20100101', 3, 32 )
INSERT INTO @Customers VALUES ('20100101', 4, 42 )
INSERT INTO @Customers VALUES ('20100101', 5, 15 )
INSERT INTO @Customers VALUES ('20100102', 1, NULL)
INSERT INTO @Customers VALUES ('20100102', 2, NULL)
INSERT INTO @Customers VALUES ('20100102', 3, 39 )
INSERT INTO @Customers VALUES ('20100102', 4, NULL)
INSERT INTO @Customers VALUES ('20100102', 5, 16 )
INSERT INTO @Customers VALUES ('20100103', 1, 13 )
INSERT INTO @Customers VALUES ('20100103', 2, 24 )
INSERT INTO @Customers VALUES ('20100103', 3, NULL)
INSERT INTO @Customers VALUES ('20100103', 4, NULL)
INSERT INTO @Customers VALUES ('20100103', 5, 21 )
INSERT INTO @Customers VALUES ('20100104', 1, 14 )
INSERT INTO @Customers VALUES ('20100104', 2, NULL)
INSERT INTO @Customers VALUES ('20100104', 3, NULL)
INSERT INTO @Customers VALUES ('20100104', 4, 65 )
INSERT INTO @Customers VALUES ('20100104', 5, 23 )
<强>查询
SELECT c.Date
, c.Customer
, Value = COALESCE(c.Value, cprevious.Value, 0)
FROM @Customers c
INNER JOIN (
SELECT c.Date
, c.Customer
, MaxDate = MAX(cdates.Date)
FROM @Customers c
LEFT OUTER JOIN (
SELECT Date
, Customer
FROM @Customers
) cdates ON cdates.Date < c.Date AND cdates.Customer = c.Customer
GROUP BY
c.Date, c.Customer
) cmax ON cmax.Date = c.Date AND cmax.Customer = c.Customer
LEFT OUTER JOIN @Customers cprevious ON cprevious.Date = cmax.MaxDate AND cprevious.Customer = cmax.Customer
ORDER BY
1, 2, 3
更新声明
UPDATE @Customers
SET Value = c2.Value
OUTPUT Inserted.*
FROM @Customers c
INNER JOIN (
SELECT c.Date
, c.Customer
, Value = COALESCE(c.Value, cprevious.Value, 0)
FROM @Customers c
INNER JOIN (
SELECT c.Date
, c.Customer
, MaxDate = MAX(cdates.Date)
FROM @Customers c
LEFT OUTER JOIN (
SELECT Date
, Customer
FROM @Customers
) cdates ON cdates.Date < c.Date AND cdates.Customer = c.Customer
GROUP BY
c.Date, c.Customer
) cmax ON cmax.Date = c.Date AND cmax.Customer = c.Customer
LEFT OUTER JOIN @Customers cprevious ON cprevious.Date = cmax.MaxDate AND cprevious.Customer = cmax.Customer
) c2 ON c2.Date = c.Date
AND c2.Customer = c.Customer
答案 4 :(得分：0)
同一个表上的左外连接如何，日期小于当前值且值为非空，按日期desc（限制1）排序，当为null时返回零？ （目前没有服务器可供测试）。除非这被视为子查询......