SQL Server：按列与另一列

Question

从下面的表中，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itemNames。相同的itemname可以出现在表格的多个位置，它们并不总是按id排序。 ItemNames是独一无二的，因为它可以通过itemname进行自我加入。

我想：

我尝试了许多使用ROW_NUMBER，按组，有等的查询，但不能得到它的权利。

有人能帮忙吗？

的样本数据：

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

下面是做到这一点的一种方式 - 使用CTE来获取所有已发送给多个电子邮件中的项目，然后加入该CTE与原始表：

;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

假设你是什么升ooking是唯一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 

产生

item1 email1 
item1 email2 
item4 email6 
item4 email7 

我们都在那里。