MySQL - Selecting row with two columns match from a table. If doesn't exist select row with one column

Question

Table:

country_id           state_id           rate
200                   100                1.1
200                   -1                 5.1

I'm trying to create a prioritized select using two combinations: 1. country_id = 200 and state_id = 100. 2. If given combination doesn't exist get country_id = 200 and state_id = -1 Note: the table is not necessarily going to have both rows, they may exist independently.

Tried to use

SELECT CASE a.rate > 0 THEN a.rate ELSE b.rate AS rate
FROM 
(
    (SELECT * FROM taxes WHERE country_id = 200 and state_id = 100) AS a,
    (SELECT * FROM taxes WHERE country_id = 200 and state_id = -1) AS b
)

another option I've tried is using UNION as subquery.

SELECT CASE b.a_rate > 0 THEN b.a_rate ELSE b.b_rate AS rate
FROM 
(
    (
       SELECT rate AS a_rate FROM taxes WHERE country_id = 200 and state_id = 100
       UNION
       SELECT rate AS b_rate FROM taxes WHERE country_id = 200 and state_id = -1
    ) AS b
)

Both work fine only if you have both rows in the table and fails if one of them is missing... any ideas?

Answer 1

The question isn't really clear, but you could do something like this

select
ifnull(
       (select rate 
       from taxes t1 
       where t1.country_id=200
       and t1.state_id = 100)
       ,(select rate
         from taxes t2
         where t2.country_id=200
         and t2.state_id = -1)
       ) as `rate`

If the first subquery returns NULL then the second subquery is used

see: http://sqlfiddle.com/#!9/802dc3/3

Answer 2

If the ID of the state -1 will be always lower than the ID for any other state I would go for the solution of adding the -1 condition to the WHERE statement, sort by the state_id field and limiting the results to take the first available:

select rate
from taxes t
where country_id = 200 and state_id in (200, -1)
order by state_id desc
limit 1;