MYSQL Count, LEFT OUTER JOIN with 0 count on a simple case

Question

Considering the two following sample tables

TABLE 1 : 'users'

ID    REGISTER_TIME    FIRSTNAME    LASTNAME    OPERATION_ID

1     1401789877       John         Doe         29
2     1401789879       Jack         Doe         29
3     1401789878       Pete         Doe         29

(Note that the register_time column is stored as INT(11))

TABLE 2 : 'calendar_days'

ID    DAY
1     2011-01-01
...   ....
n     2030-31-12

I have the following MySQL query which is working great :

SELECT d.day, COUNT(d.day)
FROM calendar_days AS d
LEFT OUTER JOIN users AS l ON DATE(FROM_UNIXTIME(l.`register_time`)) = d.day
WHERE l.`operation_id` = 29
GROUP BY d.day;

But it doesn't keep the days where count equals zero :

day;COUNT(d.day)

2014-05-07;1
2014-05-09;1
2014-05-12;11
2014-05-13;2713
2014-05-14;2631

What i want :

2014-05-07;1
**2014-05-08;0**
2014-05-09;1
2014-05-12;11
2014-05-13;2713
2014-05-14;2631

I feel like i'm close from the truth, I've tried any join possible left outer, right outer, still no empty results.

Answer 1

Your where clause turns your left join into an inner join since you use a condition on the joined table. Try

SELECT d.day, COUNT(d.day)
FROM calendar_days AS d
LEFT OUTER JOIN users AS l ON DATE(FROM_UNIXTIME(l.`register_time`)) = d.day
                           AND l.`operation_id` = 29
GROUP BY d.day;