如何计算每个客户的首次购买和退货购买？

Question

I have a table of transaction, like this:

CREATE TABLE IF NOT EXISTS `transactions` (
  `id` varchar(250) NOT NULL PRIMARY KEY,
  `date` DATE NOT NULL,
  `value` int NOT NULL,
  `customer` varchar(250) NOT NULL
);

INSERT INTO `transactions` VALUES
  ('001', '2020-12-17', 45000, 'JOHN'),
  ('002', '2020-12-25', 5000, 'MARY'),
  ('003', '2021-04-17', 1500, 'MARY'),
  ('004', '2020-06-01', 400, 'JOHN'),
  ('005', '2020-06-06', 6000, 'JOHN');

I want to calculate first purchase value and returning purchase value from each customer . I can make the select statement for first purchase value, but kind of confuse to make the returning purchase value. My query for first purchase value is like this:

SELECT customer, value as first_purchase_value, MIN(date) as first_purchase_value_date FROM orders  
GROUP BY customer

I also have one problem which I don't want to inclide minimum date into the select statement but then I can't get the first purchase value. The expected output is like this:

customer	first_purchase_value	returning_purchase_value
JOHN	45000	6000
MARY	5000	1500

I think, its can be done with LIMIT and SORT BY , but I don't know. Any idea how to solve this problem?

EDIT

Here's the fiddle: http://sqlfiddle.com/#!9/34daa6/3

Answer 1

Thanks for the Fiddle. Somehow I have a hard time using it. So I copied your code and put in my MS SQL Server. The syntax will be a bit different but it should get the job done.

Here is how I did it:

with cte_date_info as
 (
     select customer, min(date) as min_date, max(date) as max_date
     from transactions
     group by customer
 
 ),
 
 cte_first_purchase as
 (
     select customer, value as first_purchase
     from transactions
     where date IN
        (
        select min_date as date 
        from cte_date_info
        )   
 ),
 
 cte_second_purchase as
 (
     select customer, value as second_purchase
     from transactions
     where date IN
        (
        select max_date as date 
        from cte_date_info
        )   
 
 )
 
 select t1.*, t2.second_purchase
 from cte_first_purchase as t1
 left join cte_second_purchase as t2
 on t1.customer = t2.customer

I think if this is a homework, you would want to check if a customer can have more than 1 transaction in 1 day. If it's the case you will need some group by and sum(). But if only 1 transaction per day, the CTE should do the trick.

========

I think you example is inconsistent. For Mary, the subsequent purchase is 1500 which is the next purchase. But for John, his next purchase should be 400. But you put in 6000 which is the 3rd purchase.

So, do you want it to be 400, 6000, or 6400 (400 + 6000)?

======

I cannot help with the actual code as I don't have such data in my database.

But conceptually, this is what I would do,

With cte_first_purchase as
(
Calculate the first purchase of each customer
You can use rank to only keep the minimum purchase date of each person. But be depending on data structure though as each person can purchase more than once in a day.
),

With cte_subsequent_purchse as
(
calculate the subsequent purchase of each customer.

You can still use rank here but this time instead of using where = 1, then you go with where = 2. Then you will keep all the subsequent purchase, then sum the purchases if a customer buys more than once in the same day.
)

Select t1.*, t2.*
from cte_first_purchase as t1
left join cte_subsequent_purchase as t2
on t1.name = t2.name

If you can provide a code to create sample tables, I can help with more concrete codes.

Answer 2

As your comment, you can rename lastest_purchase_value as returning_... but I thing latest is more understandable

    SELECT customer,first_purchase_value,latest_purchase_value FROM 
 ( SELECT customer, value as first_purchase_value,MIN(date) FROM transactions GROUP BY customer ) r1
 JOIN 
 ( SELECT customer, value as latest_purchase_value,MAX(date) FROM transactions  GROUP BY customer) r2
 USING (customer);

Answer 3

you can try this query:

SELECT  customer
    , SUBSTRING_INDEX(GROUP_CONCAT(value order by `id`),',',1) AS first_purchase_value
    , SUBSTRING_INDEX(GROUP_CONCAT(value order by `id`),',',-1) AS returning_purchase_value
    FROM transactions
    GROUP by customer;

sample

    MariaDB [bernd]> select * from transactions;
    +-----+------------+-------+----------+
    | id  | date       | value | customer |
    +-----+------------+-------+----------+
    | 001 | 2020-12-17 | 45000 | JOHN     |
    | 002 | 2020-12-25 |  5000 | MARY     |
    | 003 | 2021-04-17 |  1500 | MARY     |
    | 004 | 2020-06-01 |   400 | JOHN     |
    | 005 | 2020-06-06 |  6000 | JOHN     |
    +-----+------------+-------+----------+
    5 rows in set (0.00 sec)
    
    MariaDB [bernd]> SELECT customer
    -> , SUBSTRING_INDEX(GROUP_CONCAT(value order by `id`),',',1) AS first_purchase_value
    -> , SUBSTRING_INDEX(GROUP_CONCAT(value order by `id`),',',-1) AS returning_purchase_value
    -> FROM transactions
    -> GROUP by customer;
+----------+----------------------+--------------------------+
| customer | first_purchase_value | returning_purchase_value |
+----------+----------------------+--------------------------+
| JOHN     | 45000                | 6000                     |
| MARY     | 5000                 | 1500                     |
+----------+----------------------+--------------------------+
2 rows in set (0.09 sec)

MariaDB [bernd]> 

NOTE: you must also add 1 index over customer plus date to have a better performance