SQL基于每行一个字段在另一列中计算重复项

Question

I am building out a customer retention report. We identify customers by their email. Here is some sample data from our table:

+----------------------------+------------------+-------------------+---------------------+------------+-------------+--------------+---------------+------------------+---------------+----------------+--------------+------------------+--+--+--+--+--+
|           Email            | BrandNewCustomer | RecurringCustomer | ReactivatedCustomer | OrderCount | TotalOrders | Date_Created | Customer_Name | Customer_Address | Customer_City | Customer_State | Customer_Zip | Customer_Country |  |  |  |  |  |
+----------------------------+------------------+-------------------+---------------------+------------+-------------+--------------+---------------+------------------+---------------+----------------+--------------+------------------+--+--+--+--+--+
| zyw@marketplace.amazon.com |                1 |                 0 |                   0 |          1 |           1 | 41:50.0      | Sha           |              990 | BRO           | NY             |          112 | US               |  |  |  |  |  |
| zyu@gmail.com              |                1 |                 0 |                   0 |          1 |           1 | 57:25.0      | Zyu           |              181 | Mia           | FL             |          330 | US               |  |  |  |  |  |
| ZyR@aol.com                |                1 |                 0 |                   0 |          1 |           1 | 10:19.0      | Day           |              581 | Myr           | SC             |          295 | US               |  |  |  |  |  |
| zyr@gmail.com              |                1 |                 0 |                   0 |          1 |           1 | 25:19.0      | Nic           |              173 | Was           | DC             |          200 | US               |  |  |  |  |  |
| zy@gmail.com               |                1 |                 0 |                   0 |          1 |           1 | 19:18.0      | Kim           |              675 | MIA           | FL             |          331 | US               |  |  |  |  |  |
| zyou@gmail.com             |                1 |                 0 |                   0 |          1 |           1 | 40:29.0      | zoe           |              160 | Mob           | AL             |          366 | US               |  |  |  |  |  |
| zyon@yahoo.com             |                1 |                 0 |                   0 |          1 |           1 | 17:21.0      | Zyo           |              84  | Sta           | CT             |          690 | US               |  |  |  |  |  |
| zyo@gmail.com              |                1 |                 0 |                   0 |          2 |           2 | 02:03.0      | Zyo           |              432 | Ell           | GA             |          302 | US               |  |  |  |  |  |
| zyo@gmail.com              |                1 |                 0 |                   0 |          1 |           2 | 12:54.0      | Zyo           |              432 | Ell           | GA             |          302 | US               |  |  |  |  |  |
| zyn@icloud.com             |                1 |                 0 |                   0 |          1 |           1 | 54:56.0      | Zyn           |              916 | Nor           | CA             |          913 | US               |  |  |  |  |  |
| zyl@gmail.com              |                0 |                 1 |                   0 |          3 |           3 | 31:27.0      | Ser           |              123 | Mia           | FL             |          331 | US               |  |  |  |  |  |
| zyk@marketplace.amazon.com |                1 |                 0 |                   0 |          1 |           1 | 44:00.0      | Myr           |              101 | MIA           | FL             |          331 | US               |  |  |  |  |  |
+----------------------------+------------------+-------------------+---------------------+------------+-------------+--------------+---------------+------------------+---------------+----------------+--------------+------------------+--+--+--+--+--+

We define our customer by email. So all orders with the same email are marked to be under one customer and then we do calculations on top of that.

Now I am trying to find out about customers whose emails have changed. So to do this we will try to line up customers by their address.

So per each row (so when separated by email), I want to have another column called something like Orders_With_Same_Address_Different_Email. How would I do that?

I have tried doing something with Dense Rank but it doesn't seem to work:

SELECT DISTINCT
Email
,BrandNewCustomer
,RecurringCustomer
,ReactivatedCustomer
,OrderCount
,TotalOrders
,Date_Created
,Customer_Name
,Customer_Address
,Customer_City
,Customer_State
,Customer_Zip
,Customer_Country
,(DENSE_RANK() over (partition by Email order by (case when email <> email then Customer_Address end)  asc) 
+DENSE_RANK() over ( partition by Email order by (case when email <> email then Customer_Address end)  desc) 
- 1) as Orders_With_Same_Name_Different_Email
--*
FROM Customers

Answer 1

Try counting the email partitioned by address, not by email:

select   Email,
         -- ...

         Orders_With_Same_Name_Different_Email = iif(
             (count(email) over (partition by Customer_Address) > 1, 
         1, 0)

from     Customers;

But this is a lesson in why you wouldn't use an email as an identifier for a client. Address is a bad idea as well. Use something that won't change. That usually means making an internal identifier, such as something that auto-increments:

alter table #customers
add customerId int identity(1,1) primary key not null

Now customerId = 1 will always refer to that particular customer.

Answer 2

You can group by customer_address and check the count. This is by the assumption that each customer has one address.

   Select * from table where 
  customer_address IN (
  Select customer_address
  From table group by customer_address
  having count(distinct customer_email) 
   >1) 

Answer 3

If I understand what you want to do, this is how I would solve it:

Note, you don't need the having clause in the CTE but depending on your data it could make it faster. (That is, if you have a large dataset.)

WITH email2addr
(
  select email, count(distinct customer_address) as addr_cnt
  from customers
  group by email
  having count(distinct customer_address) > 1
)

SELECT 
    Email
    ,BrandNewCustomer
    ,RecurringCustomer
    ,ReactivatedCustomer
    ,OrderCount
    ,TotalOrders
    ,Date_Created
    ,Customer_Name
    ,Customer_Address
    ,Customer_City
    ,Customer_State
    ,Customer_Zip
    ,Customer_Country
    CASE when coalese(email2addr.addr_cnt,1) > 1 then 'Y' ELSE 'N' END as has_more_than_1_email 
from customers
left join email2addr on customers.email = email2addr.email