SQL Server：按列与另一列分组

Question

从下面的表格中，我想获取 itemName 和电子邮件，其中相同的 itemName 已通过电子邮件发送到不同的电子邮件地址。相同的电子邮件可以接收不同的 itemName。相同的 itemname 可以出现在表格中的多个位置，它们并不总是按 id 排序。 ItemNames 是唯一的，因为可以通过 itemname 自行加入。

我试过了：

我尝试了一堆带有 row_number、group by、have 等的查询，但无法正确处理。

有人可以帮忙吗？

样本数据：

declare @t table (id int, itemname nvarchar(50), emailto nvarchar(50))

insert into @t 
values (1, 'item1', 'email1') --include 1 & 2 because same item went to different emails
       , (2, 'item1', 'email2') 
       , (3, 'item2', 'email1') --exclude because even though email1   received an email before, item2 went to a sinle email 
       , (4, 'item3', 'email3') --exclude 4, 5, 6 becuase all item3 went to the same email
       , (5, 'item3', 'email3')
       , (6, 'item3', 'email3')
       , (7, 'item4', 'email6')
       , (8, 'item4', 'email6') --include 8 & 9, only reason to exclude 7 is to get a distinct list of itemName and email pairs
       , (9, 'item4', 'email7')
       , (10, 'item3', 'email3') --exclude 10 becuase all item3 went to the same email, this is the same item from 4, 5, 6

;with expectedOutput as 
(
    select 
        t.itemname, t.emailto
    from @t t
    where 
       t.id IN (1, 2, 8, 9)
)
select *
from expectedOutput

/*
Expected output:
itemname    emailto
item1       email1
item1       email2
item4       email6
item4       email7

*/

Answer 1

这是一种方法 - 使用 CTE 获取发送到多个电子邮件的所有项目，然后将该 cte 与原始表连接：

;WITH Items AS
(
    SELECT itemname
    FROM @t
    GROUP BY itemname
    HAVING COUNT(DISTINCT emailto) > 1
)

SELECT DISTINCT t.itemname, emailto
FROM @t t
INNER JOIN Items i ON t.itemname = i.itemname

结果：

itemname    emailto
item1       email1
item1       email2
item4       email6
item4       email7

Answer 2

假设您正在寻找的是独特的电子邮件和项目对。

with expectedOutput as 
(select distinct 
  t.itemname,
  t.emailto
from @t t),
steptwo as (
    select tt.itemname, count(distinct tt.emailto) as nemails
    from expectedOutput tt
    group by tt.itemname
)
select tw.itemname,e.emailto from steptwo tw join expectedOutput e 
on tw.itemname = e.itemname 
WHERE nemails > 1

屈服

item1   email1
item1   email2
item4   email6
item4   email7

我们都去过那里。