Algorithm for available times in agenda - php

Question

I'm really stuck into this situation.
I have these 2 tables:
employee_working_schedule (stores the start and end time an employee works in a specific date)
employee_appointments
Let's suppose we have those rows stored
employee_working_schedule:
start | end
10:00 | 18:00
employee_appointments:
start | end
10:10 | 11:00
11:20 | 12:00
14:30 | 15:20
in this case I want to show that available times are:
10:00 | 10:10
11:00 | 11:20
12:00 | 14:30
15:20 | 18:00
Is there a way of doing this by SQL? I have tried to achieve with php, but no success so far.
Any help will be appreciated.

Answer 1

Here is a way to do it in pure PHP:
class TimeSpan {
function __construct($start, $end) {
$this->start = $start;
$this->end = $end;
}
function starttime() {
list($hour, $minute) = explode(":", $this->start);
return (int)$hour * 60 + (int)$minute;
}
function endtime() {
list($hour, $minute) = explode(":", $this->end);
return (int)$hour * 60 + (int)$minute;
}
}
function convert_to_time($minutes) {
$hour = (int) ($minutes / 60);
$minutes = $minutes % 60;
return str_pad($hour, 2, '0', STR_PAD_LEFT) . ':' . str_pad($minutes, 2, '0', STR_PAD_LEFT);
}
function open_times($shift, $appointments) {
$alltimes = array_fill_keys(range($shift->starttime(), $shift->endtime()), 1);
foreach ($appointments as $appt) {
$alltimes = array_diff_key($alltimes, array_fill_keys(range($appt->starttime() + 1, $appt->endtime() - 1), 1));
}
$groups = array();
$active_group = 0;
$output = array();
$output_counter = 0;
$nums = array_keys($alltimes);
foreach( $nums as $k => $num ) {
if( $k !== 0 && $nums[$k] !== $nums[$k-1]+1 ) $active_group ++;
$groups[ $active_group ][] = $num;
}
foreach( $groups as $group ) {
$first = array_shift( array_values($group) );
$output[$output_counter][] = $first;
$last = array_pop( array_values($group) );
if( $first !== $last )
$output[$output_counter][] = $last;
$output_counter++;
}
foreach ($output as &$span) {
$span[0] = convert_to_time($span[0]);
$span[1] = convert_to_time($span[1]);
}
return $output;
}
$shift = new TimeSpan("10:00", "18:00");
$appointments = array(
new TimeSpan("10:10", "11:00"),
new TimeSpan("11:20", "12:00"),
new TimeSpan("14:30", "15:20"),
);
print_r(open_times($shift, $appointments));
OUTPUT
Array
(
[0] => Array
(
[0] => 10:00
[1] => 10:10
)
[1] => Array
(
[0] => 11:00
[1] => 11:20
)
[2] => Array
(
[0] => 12:00
[1] => 14:30
)
[3] => Array
(
[0] => 15:20
[1] => 18:00
)
)

Answer 2

I can give you a solution in PHP but it would be, of course, faster if you could find something in SQL
//given you selected $employee_working_schedule
$dayStart = DateTime::createFromFormat('h:i',$employee_working_schedule['start']);
$dayEnd = DateTime::createFromFormat('h:i',$employee_working_schedule['end']);
//then assuming your query statement is named $timeQuery and use PDO
$nextFreeStart = $dayStart;
$newFreeEnd = null;
$freeTime = array();//here we will store the free time intervals
while($times = $timeQuery->fetch(PDO::FETCH_ASSOC)){
$nextFreeEnd = DateTime::createFromFormat('h:i',$times['start']);
$freeTime[] = $nextFreeEnd->diff($nextFreeStart);
$nextFreeStart = DateTime::createFromFormat('h:i',$times['end']);
}
$freeTime[] = $dayEnd->diff($nextFreeStart); //close the day

Answer 3

Strategy: Collect all startpoints, find its respective endpoint for each, then throw away rows that don't comply.
This the idea written in MySQL (untested):
SELECT eas.s AS s_start, MIN(eas.e) AS e_end
FROM ((
SELECT end AS s, start AS e
FROM employee_appointments
) UNION (
SELECT start AS s,end AS e
FROM employee_working_schedule
)) eas
WHERE end > start -- or >= if you want zero-time slots as well
GROUP BY eas.s
HAVING NOT EXISTS ( -- if appointments are disjoint, this is likely redundant
SELECT 0
FROM employee_appointments kill
WHERE s_start > kill.end AND kill.start > e_end
)