Question

我有三张桌子：

Email          | EmailID
jack@gmail.com | 100
nate@gmail.com | 101
rex@gmail.com  | 102

第二

Email          | CustomerID
jack@gmail.com | 90001
jack@gmail.com | 90002
nate@gmail.com | 90003
rex@gmail.com  | 90003

第三

CustomerID  | Purchases
90001       | 1
90002       | 2
90003       | 5

如您所见，Jack有2个与他关联的客户ID，而客户ID 90003有两个与之关联的电子邮件。我想正确地重复删除客户并给杰克1个电子邮件地址和1个客户ID（无论哪个都没关系。）

所以，理想情况下，在总结购买时我会得到以下信息：

Email          | EmailID |CustomerID   | Purchases
jack@gmail.com |  100    |90001        | 3    
nate@gmail.com |  101    |90003        | 5
nate@gmail.com |  102    |90003        | 5

Answer 1

我很困惑。得到你要求的结果将是这样的。

Name = "John"
_margin = "whatever"

Answer 2

我的解决方案，假设结果中有拼写错误

初始化：

create table #table1 (email varchar(100), emailid int)
insert into #table1 values ('jack@gmail.com', 100), ('nate@gmail.com', 101), ('rex@gmail.com',102)
create table #table2 (email varchar(100), customerid int)
insert into #table2 values ('jack@gmail.com', 90001), ('jack@gmail.com', 90002), ('nate@gmail.com',90003), ('rex@gmail.com',90003)
create table #table3 (customerid int, purchases int)
insert into #table3 values (90001, 3), (90002, 5), (90003, 5)

查询

select T1.email,
       (select max(emailid) from #table1 where email = T1.email) as 'emailid',
       (select max(customerid) from #table2 where email = T1.email) as 'customerid',
       (sum(T3.purchases)) as 'purchases' 

from #table1 T1
right join #table2 T2
on T1.email = T2.email
right join #table3 T3
on T2.customerid = T3.customerid
group by T1.email

重复数据删除多对多SQL ID

2 个答案: