SQL date difference using the same column based on conditions - mysql

Question

I'm trying to calculate the days since the last different order so for example let's say I have the following table:
cust_id|Product_id|Order_date|
1 |a |10/02/2020|
2 |b |10/01/2020|
3 |c |09/07/2020|
4 |d |09/02/2020|
1 |a |08/29/2020|
1 |f |08/02/2020|
2 |g |07/01/2020|
3 |t |06/06/2020|
4 |j |05/08/2020|
1 |w |04/20/2020|
I want to find the difference between the most recent date and the previous date that has a product ID that doesn't match the most recent product ID.
So the output should be something like this:
cust_id|latest_Product_id|time_since_last_diff_order_days|
1 |a |30 |
2 |b |92 |
3 |c |91 |
4 |d |123 |
Here's the query that I tried to use but got an error (error code 1064)
SELECT a.cust_id, a.Product_ID as latest_Product_id, DATEDIFF(MAX(a.Order_date),MAX(b.Order_date)) as time_since_last_diff_order_days
FROM database_customers.cust_orders a
INNER JOIN
database_customers.cust_orders b
on
a.cust_id = b.cust_id
WHERE a.product_id =! b.prodcut_id;
Thank you for any help!

Answer 1

It isn't pretty,, but will do the job
CREATE TABLE tab1
(`cust_id` int, `Product_id` varchar(1), `Order_date` datetime)
;
INSERT INTO tab1
(`cust_id`, `Product_id`, `Order_date`)
VALUES
(1, 'a', '2020-10-02 02:00:00'),
(2, 'b', '2020-10-01 02:00:00'),
(3, 'c', '2020-09-07 02:00:00'),
(4, 'd', '2020-09-02 02:00:00'),
(1, 'a', '2020-08-29 02:00:00'),
(1, 'f', '2020-08-02 02:00:00'),
(2, 'g', '2020-07-01 02:00:00'),
(3, 't', '2020-06-06 02:00:00'),
(4, 'j', '2020-05-08 02:00:00'),
(1, 'w', '2020-04-20 02:00:00')
;
WITH CTE AS (SELECT `cust_id`, `Product_id`,`Order_date`,ROW_NUMBER() OVER(PARTITION BY `cust_id` ORDER BY `Order_date` DESC) rn
FROM tab1)
SELECT t1.`cust_id`, t1.`Product_id`, t2.time_since_last_diff_order_days
FROM
(SELECT
`cust_id`, `Product_id`
FROM
CTE
WHERE rn = 1 ) t1
JOIN
( SELECT `cust_id`,DATEDIFF(MAX(`Order_date`), MIN(`Order_date`)) time_since_last_diff_order_days
FROM CTE WHERE rn in (1,2) GROUP BY `cust_id`) t2 ON t1.cust_id = t2.cust_id
cust_id | Product_id | time_since_last_diff_order_days
------: | :--------- | ------------------------------:
1 | a | 34
2 | b | 92
3 | c | 93
4 | d | 117
db<>fiddle here

Answer 2

I want to find the difference between the most recent date and the previous date that has a product ID that doesn't match the most recent product ID.
You can use first_value() to get the last product and then aggregate:
select cust_id, last_product_id, max(order_date),
datediff(max(order_date), max(case when product_id <> last_product_id then order_date end)) as diff_from_last_product
from (select co.*,
first_value(product_id) over (partition by cust_id order by order_date) as last_product_id
from cust_orders co
) co
group by cust_id, last_product_id;