MySQL ranking with datetime and sum - mysql

Question

I have a database where purchases are stored:
| user_id | product | price | datetime |
-----------------------------------------
| 1 | 1 | -0.75 | 2022-01-01 |
| 2 | 1 | -0.75 | 2022-01-01 |
| 3 | 2 | -0.65 | 2022-01-01 |
| 2 | 1 | -0.75 | 2022-01-01 |
| 1 | 1 | -0.75 | 2022-01-02 |
| 1 | 3 | -1.50 | 2022-01-02 |
| 1 | 2 | -0.65 | 2022-01-02 |
| 2 | 1 | -0.75 | 2022-01-02 |
| 3 | 2 | -0.65 | 2022-01-02 |
| 3 | 3 | -1.50 | 2022-01-02 |
| 3 | 3 | -1.50 | 2022-01-02 |
N.B. Time is not important in this question.
What I want is a ranking per day for each user like this for user 1:
| datetime | product1 | product2 | product3 | total | ranking |
--------------------------------------------------------------------
| 2022-01-01 | 1 | 0 | 0 | 0.75 | 2 |
| 2022-01-02 | 1 | 1 | 1 | 2.90 | 2 |
Note that the ranking is calculated for each day.
The next query gives part of the table:
SELECT
DATE(`datetime`) AS datetime,
SUM(CASE WHEN product = 1 THEN 1 ELSE 0 END) AS product1,
SUM(CASE WHEN product = 2 THEN 1 ELSE 0 END) AS product2,
SUM(CASE WHEN product = 3 THEN 1 ELSE 0 END) AS product3,
SUM(CASE WHEN product = 1 THEN 0.75 ELSE 0 END)+SUM(CASE WHEN product = 2 THEN 0.65 ELSE 0 END)+SUM(CASE WHEN product = 3 THEN 1.5 ELSE 0 END) as total,
FROM `history`
WHERE user_id=1
GROUP BY DATE(`datetime`)
My question is very similar to this one: MySQL ranking, but I can't get it exactly how I want it. It is only possible to make a ranking for the day with all users. If I add the given rank feature it will look to the table and make 2022-01-02 as the first ranking (because 2.90 is higher than 0.75). How can I make the rank look to each day?

Answer 1

The question isn't completely clear. However, what I think you're asking is how to rank the purchases for all users, by day:
history_date
user_id
DailyTotal
totalRank
2022-01-01
2
1.50
1
2022-01-01
1
0.75
2
2022-01-01
3
0.65
3
2022-01-02
3
3.65
1
2022-01-02
1
2.90
2
2022-01-02
2
0.75
3
Then display the results for a single user. So the rankings for user_id = 1 would be:
history_date
user_id
DailyTotal
totalRank
2022-01-01
1
0.75
2
2022-01-02
1
2.90
2
One way is using window functions. Aggregate the total purchases per user, by day and rank the overall total with DENSE_RANK().
Note, instead of hard coding price values, use ABS() to obtain positive numbers.
WITH cte AS (
SELECT ttl.*
, DENSE_RANK() OVER(
PARTITION BY history_date
ORDER BY DailyTotal DESC
) AS TotalRank
FROM (
SELECT user_id
, product
, price
, CAST(`datetime` AS DATE) AS history_date
, SUM( ABS(price) ) OVER(
PARTITION BY user_id, CAST(`datetime` AS DATE)
) AS DailyTotal
FROM history
WHERE product IN (1,2,3)
) ttl
)
SELECT user_id
, history_date
, SUM(CASE WHEN product = 1 THEN 1 ELSE 0 END) AS product1
, SUM(CASE WHEN product = 2 THEN 1 ELSE 0 END) AS product2
, SUM(CASE WHEN product = 3 THEN 1 ELSE 0 END) AS product3
, DailyTotal
, TotalRank
FROM cte
WHERE user_id = 1
GROUP BY user_id
, history_date
, DailyTotal
, TotalRank
;
Results:
user_id
history_date
product1
product2
product3
DailyTotal
TotalRank
1
2022-01-01
1
0
0
0.75
2
1
2022-01-02
1
1
1
2.90
2
db<>fiddle here