选择具有相同ID /电子邮件但在其他表中具有不同值的行

Question

选择具有相同ID /电子邮件但在其他表

中具有不同值的行

我有两个表：人和电子邮件，现在有相同值的邮件地址，以及具有不同值的人员/ ID。

有人能说出如何为此编写SQL查询吗？我试过但我无法弄明白。我找到了一些答案，但随后总是在同一张表中找到匹配

喜欢这个

Table_person.             ​​Table_email
1​​​                         email@persoon1
2​​​                         email@persoon2
3​​​                         email@persoon3
4​​​                         email@persoon1
5​​​                         email@persoon5
6​​​                         email@persoon2

输出应为

Table_person​​     Table_email
1​​​                email@persoon1
4​​​                email@persoon1
2​​​                email@persoon2
6​​​                email@persoon2

Answer 1

common table expression使用row_number()

;with cte as (
select *
    , rn = row_number() over (partition by email order by person_id)
from email e
)
select *
from cte
where exists (
  select 1
  from cte i
  where i.email = cte.email
    and rn > 1 
  )

或使用exists()

select *
from email e
where exists (
  select 1
  from email i
  where i.email = e.email
    and i.person_id <> e.person_id
  )

rextester演示：http://rextester.com/JHFEF82373

Answer 2

希望它会帮助你

   ;with cte(Table_person,​​Table_email)
    AS
    (
    SELECT 1​​​,'email@persoon1' UNION ALL
    SELECT 2​​​,'email@persoon2' UNION ALL
    SELECT 3​​​,'email@persoon3' UNION ALL
    SELECT 4​​​,'email@persoon1' UNION ALL
    SELECT 5​​​,'email@persoon5' UNION ALL
    SELECT 6​​​,'email@persoon2'
    )
    ,Cte2 
    AS
    (
    SELECT Table_person,​​Table_email From
    (
    Select Table_person,​​Table_email,ROW_NUMBER()OVER(Partition by Table_email Order By Table_person )Seq 
    from cte
    )dt WHERE dt.Seq>1 
    )
    ,Final
    AS
    (
    SELECT Table_person,​​Table_email From
    (
    Select Table_person,​​Table_email,ROW_NUMBER()OVER(Partition by Table_email Order By Table_email )Seq2 
    from cte
    )dt
    where dt.Seq2>1
    Union ALL
    SELECT Table_person,​​Table_email From cte2 
    )
    SELECt  Table_person,​​Table_email from Final