SQL按结果分组-Salesforce营销云

我希望找到所有重复的记录，然后从每个集合中选择所有重复项减去最旧的记录，这样我就可以删除重复项并保留一条唯一的记录

当我运行这个查询时，我得到了我想要的结果。给我留下唯一的电子邮件地址和最早的创建日期

当我向查询中添加SubscriberKey时，结果会加倍！为什么呢？我只想看到SubscriberKey绑定到我找到的子查询中日期最早的EmailAddress

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您将获得多条记录，因为您正在按SubscriberKey进行分组。您需要通过EmailAddress和CreatedDate进行匹配。尝试执行子查询并将其连接回原始表

select 
[_ListSubscribers].EmailAddress,
[_ListSubscribers].SubscriberKey,
[_ListSubscribers].CreatedDate,
from
(
SELECT 
    EmailAddress,
    MIN(CreatedDate) as CreatedDate
    FROM [_ListSubscribers]
    GROUP BY EmailAddress, SubscriberKey
    Having count(EmailAddress)>1
) SubTbl
inner join
[_ListSubscribers] on
[_ListSubscribers].EmailAddress = SubTbl.EmailAddress
and
[_ListSubscribers].CreatedDate = SubTbl.CreatedDate

我希望找到所有重复的记录，然后从每个集合中选择所有重复项减去最旧的记录，这样我就可以删除重复项并保留一条唯一的记录

使用行号：

但是，如果要删除除最新记录以外的所有记录，可以使用：

delete from _ListSubscribers
    where CreatedDate < (select max(CreatedDate)
                         from _ListSubscribers l2
                         where l2.EmailAddress = _ListSubscribers.EmailAddress
                        );

如果想要最早的记录，可以使用min而不是max来翻转逻辑

select 
[_ListSubscribers].EmailAddress,
[_ListSubscribers].SubscriberKey,
[_ListSubscribers].CreatedDate,
from
(
SELECT 
    EmailAddress,
    MIN(CreatedDate) as CreatedDate
    FROM [_ListSubscribers]
    GROUP BY EmailAddress, SubscriberKey
    Having count(EmailAddress)>1
) SubTbl
inner join
[_ListSubscribers] on
[_ListSubscribers].EmailAddress = SubTbl.EmailAddress
and
[_ListSubscribers].CreatedDate = SubTbl.CreatedDate

select l.*
from (select l.*,
             row_number() over (partition by EmailAddress order by CreatedDate desc) as seqnum
      from _ListSubscribers l
     ) l
where seqnum > 1;

delete from _ListSubscribers
    where CreatedDate < (select max(CreatedDate)
                         from _ListSubscribers l2
                         where l2.EmailAddress = _ListSubscribers.EmailAddress
                        );