Non-overlapping minutes per day - php

Question

I have been cracking my head trying to resolve this problem.
I need to know how many minutes of the day are being worked by a staff member alone in the shop.
Here is the data for daynumber = 0 (monday):
For this day, the staff member with staffid = 32 is alone from 11:00 to 11:05 in the shop.
What I have so far, is just adding all starting times, but basically what I'm thinking is, if I have any way of knowing a staff member is alone, I can calculate time between the index and the next.
for($i=0; $i<count($results); $i++){
if(isset($results[$i+1])){
if($results[$i]->starttime < $results[$i+1]->starttime)
$start = strtotime($results[$i]->starttime);
$end = strtotime($results[$i+1]->endtime);
$minutes += idate('i', $end - $start);
}
}
}
Any thoughts?
UPDATE 1:
I get to this but still no luck;
for($i=0; $i<count($results); $i++){
if(isset($results[$i+1])){
$StartDate1 = strtotime($results[$i]->starttime);
$EndDate1 = strtotime($results[$i]->endtime);
$StartDate2 = strtotime($results[$i+1]->starttime);
$EndDate2 = strtotime($results[$i+1]->endtime);
if(($StartDate1 <= $EndDate2) && ($EndDate1 >= $StartDate2)){
$StartDate1 = idate('i', $StartDate1);
$EndDate1 = idate('i', $EndDate1);
$StartDate2 = idate('i', $StartDate2);
$EndDate2 = idate('i', $EndDate2);
$a = abs($EndDate1 - $StartDate1);
$b= abs($EndDate1 - $StartDate2);
$c = abs($EndDate2 - $StartDate2);
$d = abs($EndDate2 - $StartDate1);
$minutes += min([$a,$b,$c,$d]);
}
}
}
What am I doing wrong?

Answer 1

Here's one idea, using a utility table - in this case a table of integers from 0-9.
Utility tables are frowned on by some, but I like them because they mean less typing.
You can always replace the table with a string of UNIONs.
This is for all days. I might modify it later to show how you could filter for a specific day.
SELECT * FROM ints;
+---+
| i |
+---+
| 0 |
| 1 |
| 2 |
| 3 |
| 4 |
| 5 |
| 6 |
| 7 |
| 8 |
| 9 |
+---+
SELECT SEC_TO_TIME((i4.i*1000+i3.i*100+i2.i*10+i1.i)*60) n
FROM ints i1
JOIN ints i2
JOIN ints i3
JOIN ints i4
JOIN
( SELECT daynumber
, MIN(starttime) starttime
, MAX(CASE WHEN endtime < starttime THEN SEC_TO_TIME(TIME_TO_SEC('24:00:00')+TIME_TO_SEC(endtime)) ELSE endtime END) endtime
FROM my_table
GROUP
BY daynumber
) x
ON SEC_TO_TIME((i4.i*1000+i3.i*100+i2.i*10+i1.i)*60) BETWEEN x.starttime AND x.endtime
JOIN my_table y
ON SEC_TO_TIME((i4.i*1000+i3.i*100+i2.i*10+i1.i)*60) BETWEEN y.starttime AND CASE WHEN y.endtime < y.starttime THEN SEC_TO_TIME(TIME_TO_SEC('24:00:00')+TIME_TO_SEC(y.endtime)) ELSE y.endtime END
GROUP
BY n HAVING COUNT(*) = 1;
The number of lone minutes is equal to the number of rows in this result.