display 0 to 23 hours table using php or mysql - php

Question

I have a table in mysql that has time date and in military format ( 2014-01-22 16:10:02 )
caller_data
|ID | caller_id | call_time
| 1 | 123 | 2014-01-22 16:10:02
| 2 | 123 | 2014-01-22 16:30:02
| 3 | 123 | 2014-01-22 17:10:02
| 4 | 123 | 2014-01-22 18:05:45
on the range of their time like 16:10 and 16:30 . I group them using HOUR(), the result is 16. I will count it of how many times did the caller call on the range of that hour. so that would be 2.
My query:
SELECT caller_id,
HOUR(str_to_date(date_stamp, "%Y-%m-%d %H:%i:%s")) as hourly,
count(*) as total
from caller_data
group by hourly
order by hourly;
which the result should be:
caller_id | hourly | total
123 | 16 | 2
123 | 17 | 1
123 | 18 | 1
I want to display these 3 rows on a 0-23 . I've been trying to display this with php but it display 3 times when I created a for $i <= 23 ..
for($i=0;i<=23;$i++)
{
foreach ($rec as $row):
echo $row['hourly'];
echo $row['caller_id'];
echo $row['total'];
endforeach;
}
I dont know if it's possible to display this on mysql.. can someone show how to do it with php or mysql?
| hours | caller_id | total
| 0 | 123 | 0
| 1 | 123 | 0
| 2 | 123 | 0
| 3 | 123 | 0
| 4 | 123 | 0
| 5 | 123 | 0
| 6 | 123 | 0
| 7 | 123 | 0
| 8 | 123 | 0
| 9 | 123 | 0
| 10 | 123 | 0
| 11 | 123 | 0
| 12 | 123 | 0
| 13 | 123 | 0
| 14 | 123 | 0
| 15 | 123 | 0
| 16 | 123 | 2
| 17 | 123 | 1
| 18 | 123 | 1
| 19 | 123 | 0
| 20 | 123 | 0
| 21 | 123 | 0
| 22 | 123 | 0
| 23 | 123 | 0

Answer 1

You can do this in SQL with a left outer join. You just have to type in all the hours:
select caller_id, n.n as hourly, coalesce(total, 0) as total
from (select 1 as n union all select 2 union all select 3 union all select 4 union all
select 5 union all select 6 union all select 7 union all select 8 union all
select 9 union all select 10 union all select 11 union all select 12 union all
select 13 union all select 14 union all select 15 union all select 16 union all
select 17 union all select 18 union all select 19 union all select 20 union all
select 21 union all select 22 union all select 23 union all select 24
) n left outer join
(select caller_id, HOUR(str_to_date(date_stamp, "%Y-%m-%d %H:%i:%s")) as hourly,
count(*) as total
from caller_data
group by hourly
) rep
on n.n = rep.hourly
order by hourly;