Day difference without weekends

Question

I want to count the total day difference from user input

For example when the user inputs

start_date = 2012-09-06 and end-date = 2012-09-11

For now I am using this code to find the diffeence

$count = abs(strtotime($start_date) - strtotime($end_date)); $day   = $count+86400; $total = floor($day/(60*60*24)); 

The result of total will be 6. But the problem is that I dont want to include the days at weekend (Saturday and Sunday)

2012-09-06 2012-09-07 2012-09-08 Saturday 2012-09-09 Sunday 2012-09-10 2012-09-11 

So the result will be 4

----update---

I have a table that contains date,the table name is holiday date

for example the table contains 2012-09-07

So, the total day will be 3, because it didn't count the holiday date

how do I do that to equate the date from input to date in table?

Answer 1

Very easy with my favourites: DateTime, DateInterval and DatePeriod

$start = new DateTime('2012-09-06'); $end = new DateTime('2012-09-11'); // otherwise the  end date is excluded (bug?) $end->modify('+1 day');  $interval = $end->diff($start);  // total days $days = $interval->days;  // create an iterateable period of date (P1D equates to 1 day) $period = new DatePeriod($start, new DateInterval('P1D'), $end);  // best stored as array, so you can add more than one $holidays = array('2012-09-07');  foreach($period as $dt) {     $curr = $dt->format('D');      // substract if Saturday or Sunday     if ($curr == 'Sat' || $curr == 'Sun') {         $days--;     }      // (optional) for the updated question     elseif (in_array($dt->format('Y-m-d'), $holidays)) {         $days--;     } }   echo $days; // 4