Select distinct and get sum of timestamp differences - mysql

Question

I don't know if this is possible, but it'd be really awesome. I have a table of sign-ins for people who are logging time on different projects and I need to compile a report of time logged for each project for a given time period.
My table looks something like this:
id | project | time_in | time_out | break
----------------------------------------------------------------
1 | 1 | 2014-12-07 05:00:00 | 2014-12-07 10:00:00 | 30
2 | 2 | 2014-12-07 06:00:00 | 2014-12-07 13:00:00 | 15
3 | 1 | 2014-12-07 14:00:00 | 2014-12-07 18:00:00 | 0
4 | 3 | 2014-12-07 08:30:00 | 2014-12-07 18:45:00 | 75
5 | 2 | 2014-12-07 12:00:00 | 2014-12-07 16:30:00 | 0
What I'd like to be able to do is get a report of the time logged for each project given a date range, i.e. the total time, probably in seconds, logged for each project.
time_in and time_out are fields of type TIMESTAMP; break is an integer representing the number of minutes the person was on break. I need to get the sum of time_out - time_in - break for each project, e.g. for December 7:
project | time
---------------
1 | 34200
2 | 40500
3 | 34200
This is all I have so far:
SELECT DISTINCT
`project`
FROM `sign_ins`
WHERE
`time_in` >= '2014-12-07 00:00:00' AND
`time_out` <= '2014-12-08 00:00:00';
I appreciate your help on this, SO community. You guys are so brilliant.

Answer 1

You can get the difference in seconds by converting the date/time values to Unix time stamps. Then, just aggregate the differences using sum():
SELECT project,
SUM(UNIX_TIMESTAMP(time_out) - UNIX_TIMESTAMP(time_in) - (break * 60)) as DiffSecs
FROM `sign_ins`
WHERE `time_in` >= '2014-12-07 00:00:00' AND
`time_out` <= '2014-12-08 00:00:00'
GROUP BY project;