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。同一封电子邮件可以接收不同的itemNames。相同的itemname可以出现在表中的多个位置,它们并不总是按id排序。 ItemNames是唯一的,因为它可以通过itemname自行加入。
我试过了:
我尝试了一堆查询,包括row_number,分组,等等,但无法正确查找。
有人能帮忙吗?
示例数据:
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
*/
答案 0 :(得分:3)
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与原始表格连接:
;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
答案 1 :(得分:1)
假设您要寻找的是唯一的电子邮件和物品对。
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
我们都去过那里。