Calculating working hours for each weekday between date range - mysql

Question

I have a table timeandattandance with following fields
TAId int(11) NOT NULL AUTO_INCREMENT,
PostId int(11) NOT NULL,
PositionId int(11) NOT NULL,
CreatedBy int(11) NOT NULL,
ModifiedBy int(11) NOT NULL,
CreatedDate datetime NOT NULL,
ModifiedDate datetime NOT NULL,
TimeIn datetime NOT NULL,
TimeOut datetime DEFAULT NULL,
TimeBilled tinyint(1) NOT NULL DEFAULT '0',
DeviceId varchar(50) DEFAULT NULL,
UserId int(11) DEFAULT NULL,
oldid varchar(45) DEFAULT NULL,
FromCallIn tinyint(1) DEFAULT '0'
I need to get working hours for each user for each day grouped by week ending date between a date range. that is I have been given a date range. Firstly I need to find out that timein will be between this date range then I need to get all week ending dates and then for every week I need to calculate working hours for each day.
Also while calculating the working hours I need to check if difference between timein and timeout is more than one day then these working hours will be separated for two days instead of one.
I know it's bit complex and if require more explanation please let me know.

Answer 1

Here's an example/demonstration of one possible approach:
SELECT r.dt + INTERVAL 7-DAYOFWEEK(r.dt) DAY AS week_ending
, t.userid
, r.dt
, SUM(TIMESTAMPDIFF(SECOND
,GREATEST(r.dt,t.timein)
,LEAST(r.dt+INTERVAL 1 DAY,t.timeout)
)
)/3600 AS hours_worked
FROM ( SELECT '2014-09-28' AS rb, '2014-10-11' AS re) dr
JOIN ( SELECT DATE(i.timein) AS dt FROM mytable i
UNION
SELECT DATE(o.timeout) FROM mytable o
) r
ON r.dt BETWEEN DATE(dr.rb) AND DATE(dr.re)
JOIN mytable t
ON t.timein < r.dt+INTERVAL 1 DAY
AND t.timeout > r.dt
GROUP BY t.userid, r.dt
ORDER BY week_ending, t.userid, r.dt
NOTE: The week_ending returns the date of the Saturday following (or of) the work date.
The date range is specified in the inline view dr (date range), range begin date is column rb, the range end date is column re. This example shows a two week range, starting on Sunday 2014-09-28 for two full weeks, including time worked on Saturday 2014-10-11. (The value for re could be derived as an integer number of days from rb, to get 14 full days, `rb + INTERVAL 13 DAY)
Only hours worked on these dates, or on days between these dates, are reported. A given userid that did not have any work "time" on a given date will not have a row returned for that date. (The query could be easily tweaked to return rows for all employees for all dates in the range, and returning zeros.)
Absent a "cal" calendar table that contains all date values, we can get a distinct list of date values in the range from the table itself; this could be relatively expensive operation for a large number or rows in the table. This won't return date values that don't appear in the timein or timeout columns. (If there's a work period of over 24 hours, e.g. starting on Monday and ending on Wednesday. and there are no other rows that have a timein or timeout on Wednesday, the 24 hours worked that day will be omitted... that's likely an extreme corner case. Having a distinct list of all possible date value available in a calendar table avoids that problem.)
FOLLOWUP
From your comment it sounds like you need a cross product between a distinct list of days, and a distinct list of users, then an outer join to the working hours table.
JOIN ( distinct_list_of_dates_r ) r
ON ( r_in_specified_week )
CROSS
JOIN ( distinct_list_of_userid_u ) u
LEFT
JOIN mytable t
ON ( t_userid_matches_u )
AND ( t_times_matches_r )
And then GROUP BY the userid from u, rather than from t. You'll likely want to replace a NULL value returned from hours_worked with a 0. The MySQL IFNULL function is convenient for that, IFNULL(expr,0) is shorthand for CASE WHEN expr IS NULL THEN 0 ELSE expr END. If you want total hours for the entire week, rather than by individual days, then do the GROUP BY on the week_ending expression, rather than on the individual date.