MySQL: Rank the total total amount with COUNT() - mysql

Question

Edit: I want to rank with COUNT()
I've made a View with every order a customer made. In the next step I wrote a query to calculate the total amount a customer purchased.
Now I want to rank the customers, based on their total purchase.
I wrote this query:
SELECT u.m_name, SUM(u.num * u.price) AS total,
(SELECT COUNT(*)
FROM v_sales AS x
WHERE x.m_id = u.m_id
AND (SELECT SUM(s1.num * s1.price) FROM v_sales AS s1 WHERE s1.m_id = x.m_id)
>
(SELECT SUM(s2.num * s2.price) FROM v_sales AS s2 WHERE s2.m_id = x.m_id)
) + 1 AS Rank
FROM v_sales AS u
GROUP BY u.m_id;
But the results are not the expected ones:
# m_name total Rank
川島智弘 2620 1
河田英毅 0 1
山田忠明 15420 1
永峰弘万 500 1
永山智広 380 1
I need the following output:
# m_name total Rank
川島智弘 2620 2
河田英毅 0 5
山田忠明 15420 1
永峰弘万 500 3
永山智広 380 4
Has someone an idea what I did wrong? It would be also helpful if someone could explain why my query doesn't work.
Here is a Fiddle
Thank you

Answer 1

You can use the RANK function in MariaDB 10.4.
SELECT m_name, SUM(num * price) AS total,
RANK() OVER(ORDER BY SUM(num * price) DESC)
FROM v_sales
GROUP BY m_id;
Fiddle
No window function:
SELECT t1.m_name,MAX(t1.total),COUNT(t2.m_name)+1 as RANK
FROM
(SELECT m_name, SUM(num * price) AS total FROM
v_sales
GROUP BY m_id) t1
LEFT JOIN
(SELECT m_name, SUM(num * price) AS total FROM
v_sales
GROUP BY m_id) t2
ON t1.total<t2.total
GROUP BY t1.m_name
ORDER BY 3
Fiddle