Calculate hours within a date range and given time limit - php

Question

Lets say I have the following daterange:
2014-01-10 11:00 - 2014-01-13 15:00
I also know that I can only count the hours between 09:00 - 16:00, how do I achieve this with PHP?
The above daterange should give me 25 hours (5 + 7 + 7 + 6), not sure how such PHP-function should look, though.

Answer 1

Completely untested, without any varatny
<?php
$time1 = new DateTime("11:00");
$time2 = new DateTime("15:00");
$date1 = new DateTime("2014-01-10");
$date2 = new DateTime("2014-01-13");
$start = new DateTime("9:00");
$end = new DateTime("16:00");
$maxperday = $end->diff($start)->format("%h");
$full_day_hours = ($date2->diff($date1)->format("%a")-1)*$maxperday;
$first_day_hours = min($maxperday,max(0,$end->diff($time1)->format("%h")));
$last_day_hours = min($maxperday,max(0,$time2->diff($start)->format("%h")));
$hours = $first_day_hours + $full_day_hours + $last_day_hours;
?>
EDIT: Tested now, corrected two errors, seems to work now.

Answer 2

I have made a solution now, but not too happy with my calculation if it is just one day. Here is the code:
$startDate = strtotime("2014-01-22 15:00");
$endDate = strtotime("2014-01-24 18:00");
$time1 = new DateTime(date("H:i", $startDate));
$time2 = new DateTime(date("H:i", $endDate));
$date1 = new DateTime(date("Y-m-d", $startDate));
$date2 = new DateTime(date("Y-m-d", $endDate));
$start = new DateTime("11:00");
$end = new DateTime("18:00");
$maxPerDay = $end->diff($start)->format("%h") != 0 ? $end->diff($start)->format("%h") : 24;
$days = $date2->diff($date1)->format("%a");
$hours = 0;
//Is it just one day and is time intersecting?
if($days == 0 && ($start <= $time2 && $time1 <= $end)){
$hours = $maxPerDay;
if($start < $time1 || $end > $time2){
if($time1 < $start){
$hours = $maxPerDay - $time2->diff($end)->format("%h");
}else if($time2 > $end){
$hours = $maxPerDay - $time1->diff($start)->format("%h");
}else{
$hours = $time1->diff($time2)->format("%h");
}
}
}else if($days > 0){
$firstDay = 0;
$lastDay = 0;
if($time1 < $end){
$firstDay = $time1 < $start ? $maxPerDay : $maxPerDay - $time1->diff($start)->format("%h");
}
if($time2 > $start){
$lastDay = $time2 > $end ? $maxPerDay : $maxPerDay - $time2->diff($end)->format("%h");
}
$hours = $lastDay + $firstDay + ($days - 1) * $maxPerDay;
}
echo $hours;