MYSQL query that returns a default row if no other row is matched

Question

I have an example table "cities" like this:

id   |   city
----------------
1      London
2      Liverpool
3      Birmingham
4      (Other)

Can I write a query that returns (Other) if I provide a non-existing city... such as:

SELECT * FROM cities WHERE city = "London"      --> {1, London}
SELECT * FROM cities WHERE city = "Manchester"  --> {4, (Other) }

I understand that the second SELECT-statement will return nothing. But how can I modify it to return the "(Other)" row? Conceptual code:

SELECT * FROM cities WHERE city = "Manchester" OTHERWISE WHERE city = "(Other)"

Answer 1

One method is:

select c.*
from cities c
where c.city in (?, '(Other)')
order by c.city = '(Other)'
limit 1;

This retrieves the two rows possible rows that might match (? is a placeholder for the name you want). If there are two rows, then then one that is not "other" is chosen.

Answer 2

Consider:

select * 
from cities 
where 
    city = 'Manchester'
    or (city = '(Other)' and not exists (select 1 from cities where city = 'Manchester'))