L'ho capito modificando la query fornita nella soluzione per ottenere tutte le date.
La query seguente restituisce tutte le date e conta gli ID se esistono record:
select d.date, count(v.id) from
(select adddate('1970-01-01',t4.i*10000 + t3.i*1000 + t2.i*100 + t1.i*10 + t0.i) date from
(select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t0,
(select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t1,
(select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t2,
(select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t3,
(select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t4) d
left join visitors v on d.date = v.date
where d.date between '2016-06-01' and '2016-06-30'
group by d.date
order by d.date
La cortesia per ottenere l'intervallo di date va a @mark-bannister e un semplice join sulla query che corrisponde ai risultati e l'ordinamento ottiene la soluzione.
