get total working hour based of day of the week - mysql

Question

I want to check if employee's daily working hour is less than total working hour assigned.
For example employee #1 are assigned to work :
Day | Total working hours
mon | 10
tue | 10
wed | 10
thu | 10
fri | 10
sat | 0
sun | 0
I have total working hour of employee #1 as below :
Start | End | Total Hours Worked
2018-07-02 00:28:29 | 2018-07-02 04:12:17 | 3.72
2018-07-05 00:26:20 | 2018-07-05 05:03:23 | 4.62
2018-07-12 00:27:35 | 2018-07-12 10:21:08 | 9.88
The problem I'm facing is to know which day of the week to check the total working hour based on date. e.g. 2018-07-02 is monday so the total working hour employee should work is 10 hours however the total hour worked by the employee is only 3.72. Hence for 2018-07-02 employee has not fullfill total working hour assigned.
The expected query result should return :
Short Working Hour Date
2018-07-02
2018-07-05
2018-07-12
Appreciate any help from you guys. Thanks 🙏🏻

Answer 1

I found a way to do it refering the manual. But it looks messy.
SELECT
start,
end,
total_hours_worked
FROM mytable
GROUP BY DATE(start), DATE(end)
having CASE
WHEN DAYOFWEEK(MIN(start))=1 and total_hours_worked < 0 THEN 1
WHEN DAYOFWEEK(MIN(start))=2 and total_hours_worked < 10 THEN 1
WHEN DAYOFWEEK(MIN(start))=3 and total_hours_worked < 10 THEN 1
WHEN DAYOFWEEK(MIN(start))=4 and total_hours_worked < 10 THEN 1
WHEN DAYOFWEEK(MIN(start))=5 and total_hours_worked < 10 THEN 1
WHEN DAYOFWEEK(MIN(start))=6 and total_hours_worked < 10 THEN 1
WHEN DAYOFWEEK(MIN(start))=7 and total_hours_worked < 0 THEN 1
ELSE 0
END;

Answer 2

I don't understand exactly what you're after, so here's an incomplete solution...
DROP TABLE employee_assignments;
CREATE TABLE employee_assignments
(day CHAR(3) NOT NULL PRIMARY KEY
,total_working_hours INT NOT NULL DEFAULT 0
);
INSERT INTO employee_assignments VALUES
('mon',10),
('tue',10),
('wed',10),
('thu',10),
('fri',10),
('sat',0),
('sun',0);
DROP TABLE timesheet;
CREATE TABLE timesheet
(start DATETIME NOT NULL PRIMARY KEY
,end DATETIME NOT NULL
);
INSERT INTO timesheet VALUES
('2018-07-02 00:28:29','2018-07-02 04:12:17'),
('2018-07-05 00:26:20','2018-07-05 05:03:23'),
('2018-07-12 00:27:35','2018-07-12 10:21:08');
SELECT t.*
, (TIME_TO_SEC(t.end)-TIME_TO_SEC(t.start))/3600 n
, a.*
FROM timesheet t
JOIN employee_assignments a
ON a.day = DATE_FORMAT(t.start,'%a');
+---------------------+---------------------+--------+-----+---------------------+
| start | end | n | day | total_working_hours |
+---------------------+---------------------+--------+-----+---------------------+
| 2018-07-02 00:28:29 | 2018-07-02 04:12:17 | 3.7300 | mon | 10 |
| 2018-07-05 00:26:20 | 2018-07-05 05:03:23 | 4.6175 | thu | 10 |
| 2018-07-12 00:27:35 | 2018-07-12 10:21:08 | 9.8925 | thu | 10 |
+---------------------+---------------------+--------+-----+---------------------+