MySQL SELECT MIN price with priority of field - mysql

Question

uI have a database with 3 price fields.
Now I want to query the smallest price in the following order
If MIN price 1 is greater 0 then price 1
ELSE If MIN price 2 is greater 0 then price 2
ELSE If MIN price 3 is greater 0 then price 3
But I always get the lowest price, no matter in which field this is
SELECT f.user,
(SELECT
(
CASE
WHEN MIN(s.price2) > '0' THEN CONCAT(MIN(s.price2), '*Week*')
WHEN MIN(s.price1) > '0' THEN CONCAT(MIN(s.price1), '*Week*')
WHEN MIN(s.price3) > '0' THEN CONCAT(MIN(s.price3), '*Week*')
ELSE 0
END
)
FROM prices s WHERE f.user = s.user
) as price FROM table f WHERE f.user = 84
+------+--------+--------+--------+--------+
| id | user | price1 | price2 | price3 |
+------+--------+--------+--------+--------+
| 1 | 84 | 785 | 0 | 0 |
| 2 | 84 | 550 | 200 | 100 |
| 3 | 84 | 0 | 90 | 45 |
| 4 | 84 | 0 | 0 | 80 |
| 5 | 84 | 0 | 300 | 80 |
| 6 | 84 | 100 | 95 | 0 |
+------+--------+--------+--------+--------+
What i want: Priority has MIN price2, then MIN price1, then MIN price4.
+------+-------+-------+
| id | user | price |
+------+-------+-------+
| 1 | 84 | 90 |
+------+-------+-------+

Answer 1

I am still not sure I understand your question correctly. You take the minumum prices:
MIN(price1) = 100
MIN(price2) = 90
MIN(price3) = 45
SUM(price2) has highest priority. As it is greater than zero we return this. (Otherwise we'd check SUM(price1) and if that were zero or less too, we'd check SUM(price3).)
select
case
when sum(price2) > 0 then sum(price2)
when sum(price1) > 0 then sum(price1) end
when sum(price3) > 0 then sum(price3) end
else 0
end as price
from prices
where user = 84;