Count distinct values in MySQL without folding NULL values

Question

I have table a with two fields: id (PK) and f.

Consider following records:

id | f
1  | NULL
2  | 'foo'
3  | 'bar'
4  | NULL
5  | 'foo'
6  | 'baz'

I want to retrieve and count all the records having distinct f values including every record WHERE f IS NULL. Given this criteria, the query should return every record except #5, because the same value is already included in the set, and the total count would be 5.

The query I'm using to retrieve all records looks like this:

SELECT CASE WHEN EXISTS (SELECT id FROM a a2 WHERE a2.f = a.f AND a.id < a2.id) THEN 1 END AS not_distinct FROM a HAVING not_distinct IS NULL

If this query could be improved, I'd welcome any feedback. Anyway, the main problem is counting. Obviously adding a COUNT(*) will not help here and I'm totally lost how to count the records after the filtering.

Answer 1

There's a pretty simple approach that might work for you:

select count(distinct ifnull(f, id))
from a

Note that this query assumes that f values are never id values, and based on sample data and experience this is reasonable.

Edited:

I thought about it and there's an even simpler approach:

select count(distinct f) + sum(f is null) from a;

which you can see running on sqlfiddle

This works because distinct throws away nulls, and sum(condition) counts the number of times condition is true because in mysql true is 1 and false is 0.