Get total hours with PHP & MySQL - php

Question

I have the following table
id | user_id | date | status
1 | 53 | 2018-09-18 06:59:54 | 1
2 | 62 | 2018-09-18 07:00:16 | 1
3 | 53 | 2018-09-18 09:34:12 | 2
4 | 53 | 2018-09-18 12:16:27 | 1
5 | 53 | 2018-09-18 18:03:19 | 2
6 | 62 | 2018-09-18 18:17:41 | 2
I would like to get the total working hours (from date range) and group them by user_id
UPDATE
The system does not "require" a check-out so if there is only one value can we set a default check out time lets say 19:00:00? IF not I can check every day at 21:00:00 if there is not a checkout time to manually insert it at 19:00:00
UPDATE 2
I have added a new field in the table "status" so the very first check-in of the date the status = 1 and every 2nd check-in the status = 2
So if a user check-ins for the 3rd time during the day the status will be 1 again etc.
I hope this will make things easier
Thanks

Answer 1

In case of multiple check-in and check-out happening within a day, for a user:
Utilizing Correlated Subquery, we can find corresponding "checkout_time" for every "checkin_time".
Also, note the usage of Ifnull(), Timestamp() functions etc, to consider default "checkout_time" as 19:00:00, in case of no corresponding entry.
Then, considering this enhanced data-set as Derived Table, we group the data-set based on the user_id and date. Date (yyyy-mm-dd) can be determined using Date() function.
Eventually, use Timestampdiff() function with Sum aggregation, to determine the total work seconds for a user_id at a particular date.
You can easily convert these total seconds to hours (either in your application code, or at the query itself (divide seconds by 3600).
The reason I have preferred to compute using seconds, as Timestampdiff() function returns integer only. So there may be truncation errors, in case of multiple checkin/checkout(s).
Use the following query (replace your_table with your actual table name):
SELECT inner_nest.user_id,
DATE(inner_nest.checkin_time) AS work_date,
SUM(TIMESTAMPDIFF(SECOND,
inner_nest.checkin_time,
inner_nest.checkout_time)) AS total_work_seconds
FROM
(
SELECT t1.user_id,
t1.date as checkin_time,
t1.status,
IFNULL( (
SELECT t2.date
FROM your_table AS t2
WHERE t2.user_id = t1.user_id
AND t2.status = 2
AND t2.date > t1.date
AND DATE(t2.date) = DATE(t1.date)
ORDER BY t2.date ASC LIMIT 1
),
TIMESTAMP(DATE(t1.date),'19:00:00')
) AS checkout_time
FROM `your_table` AS t1
WHERE t1.status = 1
) AS inner_nest
GROUP BY inner_nest.user_id, DATE(inner_nest.checkin_time)
Additional: Following solution will work for the case when there is a single check-in, and corresponding check-out on the same date.
You first need to group the dataset based on the user_id and date. Date (yyyy-mm-dd) can be determined using Date() function.
Now use aggregation functions like Min() and Max() to find the starting and closing time for a user_id at a particular date.
Eventually, use Timestampdiff() function to determine the working hours for a user_id at a particular date (difference between the closing and starting time)
Try the following query (replace your_table with your actual table name):
SELECT user_id,
DATE(`date`) AS working_date,
TIMESTAMPDIFF(HOUR, MIN(`date`), MAX(`date`)) AS working_hours
FROM your_table
GROUP BY
user_id,
DATE(`date`)

Answer 2

Use TIMESTAMPDIFF function
the query more like :
SELECT t1.user_id, TIMESTAMPDIFF(HOUR,t1.date,t2.date) as difference
FROM your_table t1
INNER JOIN your_table t2 on t1.user_id = t2.user_id
Group By t1.user_id
You can see this as preference TimeStampDiff