Sql server 用于合并客户帐户的SQL脚本

我已经想出了一个脚本来选择“主”帐户和“从”帐户。公司名称和邮政编码完全匹配的地方。它认为最近更新的帐户是主帐户

select
    m.ev870_acct_code, m.ev870_company_name, m.ev870_postal_code, m.ev870_iacvb_code,
    s.ev870_acct_code, s.ev870_company_name, s.ev870_postal_code, s.ev870_iacvb_code
from
    ev870_acct_master m
inner join
    ev870_acct_master s
on
    m.ev870_company_name = s.ev870_company_name
and m.ev870_postal_code = s.ev870_postal_code
and m.ev870_upd_stamp > s.ev870_upd_stamp
where
    m.ev870_class = 'o'
and s.ev870_class = 'o'
and m.ev870_status != '0'
and s.ev870_status != '0'
and (m.ev870_iacvb_code = s.ev870_iacvb_code or isnull(m.ev870_iacvb_code,'') = '' or isnull(s.ev870_iacvb_code,'') = '')
and s.ev870_company_name like '%council%'
order by
    m.ev870_upd_stamp desc

当前脚本的问题在于，它可能会确定：

• 帐户1是主帐户，存在重复的从属帐户2
• 帐户1是主帐户，存在重复的从属帐户3
• 帐户2是主帐户，存在重复的从属帐户3

正如您所见，每个步骤的结果都会影响下一步。你能推荐一个更聪明的查询吗

编辑解决方案：

select
    m.ev870_acct_code, m.ev870_company_name, m.ev870_postal_code, m.ev870_iacvb_code,
    s.ev870_acct_code, s.ev870_company_name, s.ev870_postal_code, s.ev870_iacvb_code
from
    ev870_acct_master s
inner join 
    (
    select 
        ev870_acct_code, ev870_company_name, ev870_postal_code, ev870_iacvb_code, ev870_upd_stamp
        ,row_number() over (partition by ev870_company_name, ev870_postal_code, ev870_iacvb_code order by ev870_upd_stamp desc) as howRecent
    from 
        ev870_acct_master
    where
        ev870_class = 'o'
    and ev870_status != '0'
    and ev870_postal_code != ''
    and ev870_company_name like 'A%'
    ) m 
on  
    m.ev870_company_name = s.ev870_company_name
and m.ev870_postal_code = s.ev870_postal_code
and m.ev870_upd_stamp > s.ev870_upd_stamp
where
    m.howRecent = 1
and m.ev870_iacvb_code = s.ev870_iacvb_code
and s.ev870_class = 'o'
and s.ev870_status != '0'

---

以下是您的评论：

@kristof我在我的应用程序中识别重复帐户 然后将合并到单个帐户中

您可以使用类似于以下内容的代码：

declare @dupExample table (
    id int identity(1,1)
    ,name varchar(50)
    ,postal varchar(50)
    ,lastUpdated datetime
)

insert into @dupExample(name, postal, lastUpdated)
values 
    ('a','pc1','20120101')
    ,('a','pc1','20120501')
    ,('a','pc1','20120601')
    ,('a','pc1','20120701')
    ,('a','pc1','20120201')
    ,('b','pc2','20120102')
    ,('b','pc2','20120202')
    ,('b','pc2','20120302')
    ,('b','pc2','20120302')
    ,('c','pc2','20120302')
    ,('d','pc2','20120302')
    ,('d','pc2','20120302')


select * from @dupExample 


/*
    to see duplicates along with how recent they are
*/
select 
    *
    ,row_number() over (partition by name, postal order by lastUpdated) as howRecent
from 
    @dupExample     

/*
    delete duplicates leaving only most recent record based on date updated
    WARNING only one record will be left for each dup even if there are multiple records 
    updated on the same date (see b, d examples)
*/
delete de
from    @dupExample de
    inner join 
    ( select 
        id
        ,row_number() over (partition by name, postal order by lastUpdated desc) as howRecent
        from 
            @dupExample     
    ) der on de.id = der.id
where
    der.howRecent > 1   

/*
    after delete
*/      
select * from @dupExample   

---

如果您有更新日期相同的重复项，您可以在“排序依据”部分中添加一些额外的条件，以指定在这种情况下要删除的记录-但希望它能为您提供一个良好的起点。

您使用的是什么版本的SQL Server？@Lee您能解释一下您试图通过此查询完成的任务吗？SQL服务器2005@kristof我在我的应用程序中识别重复的帐户，然后将其合并为一个帐户。很好-谢谢，这不是我想要做的-但基本上给了我解决方案（请参阅我问题中的注释）。