SQL server 2005-3表难题
SQL Server 2005在2000兼容模式下运行 我有三张表:银行、例外和客户;客户可以与任何银行进行交易,例外情况表中与其关联的银行除外。基本上,它们看起来像下面的模式:SQL server 2005-3表难题,sql,sql-server-2005,Sql,Sql Server 2005,SQL Server 2005在2000兼容模式下运行 我有三张表:银行、例外和客户;客户可以与任何银行进行交易,例外情况表中与其关联的银行除外。基本上,它们看起来像下面的模式: Banks Exceptions Clients ----- ---------- ------- bkID bkID clID clID 我的问题是:我如何才能找到
Banks Exceptions Clients
----- ---------- -------
bkID bkID
clID clID
我的问题是:我如何才能找到所有客户,他们最多有两家银行允许与他们交易,即不在例外表上,并且这两家银行中的一家是特定的,对所有客户来说都是相同的
另一种重新表述问题的方式是:我如何找到所有拥有一家特定银行的客户?我们可以选择JPM作为例子,最多选择一家其他银行,允许他们与之交易
到目前为止,我已经创建了一个函数来计算给定一个clID有多少银行可用,但是我真的不知道如何添加所有至少有共同JPM的客户的条件
谢谢大家
PS:功能代码
ALTER FUNCTION [dbo].[fnGetNbAvailableBanks](@clientID varchar(10))
RETURNS INTEGER
AS
BEGIN
declare @intReturn integer
set @intReturn = (select
count(*) numBanks
from
Banks fxb
left outer join Exceptions bx
on bx.clID= @clientID and fxb.bkID = bx.bkID
where
bx.bkID is null
and isnull(fxb.bObsolete, 0) = 0)
RETURN @intReturn
END
到目前为止,我正在尝试运行的查询:
select *
from
(select
clID,
dbo.fnGetNbAvailableBanks(clID) cnt
from
Clients) t
where t.cnt <= 2
正如您所看到的,因为我只返回带有的客户端,这应该可以工作:
select clID from
(select distinct t1.bkID, t2.clID from Clients t1, Banks t2
where t2.bkID not in
(select bkID from Exceptions t3 where t1.clID = t3.clID ) ) as t1
where clID not in (select clID from Exceptions where bkID = %yourspecific bankid%)
group by clID
having count(*) <= 2
您可以使用哈希表,而不是选择括号中的station,以加快查询速度。这应该可以:
select clID from
(select distinct t1.bkID, t2.clID from Clients t1, Banks t2
where t2.bkID not in
(select bkID from Exceptions t3 where t1.clID = t3.clID ) ) as t1
where clID not in (select clID from Exceptions where bkID = %yourspecific bankid%)
group by clID
having count(*) <= 2
您可以使用哈希表,而不是选择括号中的station,以加快查询速度。我将函数更改为使用EXISTS,如下所示:
ALTER FUNCTION [dbo].[FNGETNBAVAILABLEBANKS](@clID VARCHAR(10))
returns INTEGER
AS
BEGIN
DECLARE @intReturn INTEGER
SET @intReturn = (SELECT COUNT(*) numBanks
FROM (SELECT fxb.bkID
FROM Banks fxb
LEFT OUTER JOIN Exceptions bx
ON bx.clID = @clID
AND fxb.bkID = bx.bkID
WHERE bx.bkID IS NULL
AND ISNULL(fxb.bObsolete, 0) = 0
AND EXISTS (SELECT fxb.bkID
FROM Banks fxb
LEFT OUTER JOIN
Exceptions bx
ON bx.clID = @clID
AND fxb.bkID = bx.bkID
WHERE bx.bkID IS NULL
AND fxb.bkID = 'JPM')) a)
RETURN @intReturn
END
select *
from
(select
clID,
dbo.fnGetNbAvailableBanks(clID) cnt
from
Clients) t
where t.cnt <= 2 and t.cnt > 0
查询如下:
ALTER FUNCTION [dbo].[FNGETNBAVAILABLEBANKS](@clID VARCHAR(10))
returns INTEGER
AS
BEGIN
DECLARE @intReturn INTEGER
SET @intReturn = (SELECT COUNT(*) numBanks
FROM (SELECT fxb.bkID
FROM Banks fxb
LEFT OUTER JOIN Exceptions bx
ON bx.clID = @clID
AND fxb.bkID = bx.bkID
WHERE bx.bkID IS NULL
AND ISNULL(fxb.bObsolete, 0) = 0
AND EXISTS (SELECT fxb.bkID
FROM Banks fxb
LEFT OUTER JOIN
Exceptions bx
ON bx.clID = @clID
AND fxb.bkID = bx.bkID
WHERE bx.bkID IS NULL
AND fxb.bkID = 'JPM')) a)
RETURN @intReturn
END
select *
from
(select
clID,
dbo.fnGetNbAvailableBanks(clID) cnt
from
Clients) t
where t.cnt <= 2 and t.cnt > 0
很抱歉没有缩进,我不知道为什么它们没有出现。。。我当然缩进了代码…我将函数改为使用EXISTS,如下所示:
ALTER FUNCTION [dbo].[FNGETNBAVAILABLEBANKS](@clID VARCHAR(10))
returns INTEGER
AS
BEGIN
DECLARE @intReturn INTEGER
SET @intReturn = (SELECT COUNT(*) numBanks
FROM (SELECT fxb.bkID
FROM Banks fxb
LEFT OUTER JOIN Exceptions bx
ON bx.clID = @clID
AND fxb.bkID = bx.bkID
WHERE bx.bkID IS NULL
AND ISNULL(fxb.bObsolete, 0) = 0
AND EXISTS (SELECT fxb.bkID
FROM Banks fxb
LEFT OUTER JOIN
Exceptions bx
ON bx.clID = @clID
AND fxb.bkID = bx.bkID
WHERE bx.bkID IS NULL
AND fxb.bkID = 'JPM')) a)
RETURN @intReturn
END
select *
from
(select
clID,
dbo.fnGetNbAvailableBanks(clID) cnt
from
Clients) t
where t.cnt <= 2 and t.cnt > 0
查询如下:
ALTER FUNCTION [dbo].[FNGETNBAVAILABLEBANKS](@clID VARCHAR(10))
returns INTEGER
AS
BEGIN
DECLARE @intReturn INTEGER
SET @intReturn = (SELECT COUNT(*) numBanks
FROM (SELECT fxb.bkID
FROM Banks fxb
LEFT OUTER JOIN Exceptions bx
ON bx.clID = @clID
AND fxb.bkID = bx.bkID
WHERE bx.bkID IS NULL
AND ISNULL(fxb.bObsolete, 0) = 0
AND EXISTS (SELECT fxb.bkID
FROM Banks fxb
LEFT OUTER JOIN
Exceptions bx
ON bx.clID = @clID
AND fxb.bkID = bx.bkID
WHERE bx.bkID IS NULL
AND fxb.bkID = 'JPM')) a)
RETURN @intReturn
END
select *
from
(select
clID,
dbo.fnGetNbAvailableBanks(clID) cnt
from
Clients) t
where t.cnt <= 2 and t.cnt > 0
很抱歉没有缩进,我不知道为什么它们没有出现。。。我当然缩进了代码…发布你所有的代码,包括你的函数代码,可能会有帮助,你离正确的解决方案不远。你的设计有缺陷:例外表应该允许使用TradeWith表。谢谢Alex,但是正如你所能想象的那样,现在不由我来更改设计。这是一个遗留系统,需要几个月的回归测试才能改变任何东西。约束通常会培养创造力,不是吗?发布您拥有的任何代码,包括函数代码,可能会有所帮助,而且您离正确的解决方案不远。您的设计存在缺陷:应允许例外表与表进行交易。谢谢Alex,但正如您所想象的,现在不由我来更改设计。这是一个遗留系统,需要几个月的回归测试才能改变任何东西。约束通常会培养创造力,没有瑕疵,优雅,速度相当快,尤其是与我天真的版本相比。谢谢你!完美无瑕、优雅且速度相当快,尤其是与我天真的版本相比。谢谢你!