php – Day difference without weekends-ThrowExceptions

Exception or error:

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?

How to solve:

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

Answer:

In my case I needed the same answer as OP, but wanted something a little smaller. @Bojan’s answer worked, but I didn’t like that it doesn’t work with DateTime objects, required using timestamps, and was comparing against strings instead of the actual objects themselves (which feels hacky)… Here’s a revised version of his answer.

function getWeekdayDifference(\DateTime $startDate, \DateTime $endDate)
{
    $days = 0;

    while($startDate->diff($endDate)->days > 0) {
        $days += $startDate->format('N') < 6 ? 1 : 0;
        $startDate = $startDate->add(new \DateInterval("P1D"));
    }

    return $days;
}

Per @xzdead’s comment if you’d like this to be inclusive of the start and end date:

function getWeekdayDifference(\DateTime $startDate, \DateTime $endDate)
{
    $isWeekday = function (\DateTime $date) {
        return $date->format('N') < 6;
    };

    $days = $isWeekday($endDate) ? 1 : 0;

    while($startDate->diff($endDate)->days > 0) {
        $days += $isWeekday($startDate) ? 1 : 0;
        $startDate = $startDate->add(new \DateInterval("P1D"));
    }

    return $days;
}

Answer:

use DateTime:

$datetime1 = new DateTime('2012-09-06');
$datetime2 = new DateTime('2012-09-11');
$interval = $datetime1->diff($datetime2);
$woweekends = 0;
for($i=0; $i<=$interval->d; $i++){
    $datetime1->modify('+1 day');
    $weekday = $datetime1->format('w');

    if($weekday !== "0" && $weekday !== "6"){ // 0 for Sunday and 6 for Saturday
        $woweekends++;  
    }

}

echo $woweekends." days without weekend";

// 4 days without weekends

Answer:

The easiest and fastest way to get difference without weekends is by using Carbon library.

Here’s an example how to use it:

<?php

$from = Carbon\Carbon::parse('2016-05-21 22:00:00');
$to = Carbon\Carbon::parse('2016-05-21 22:00:00');
echo $to->diffInWeekdays($from);

Answer:

date(‘N’) gets the day of the week (1 – Monday, 7 – Sunday)

$start = strtotime('2012-08-06');
$end = strtotime('2012-09-06');

$count = 0;

while(date('Y-m-d', $start) < date('Y-m-d', $end)){
  $count += date('N', $start) < 6 ? 1 : 0;
  $start = strtotime("+1 day", $start);
}

echo $count;

Answer:

Have a look at this post:
Calculate business days

(In your case, you could leave out the ‘holidays’ part since you’re after working/business days only)

<?php
//The function returns the no. of business days between two dates
function getWorkingDays($startDate,$endDate){
    // do strtotime calculations just once
    $endDate = strtotime($endDate);
    $startDate = strtotime($startDate);


    //The total number of days between the two dates. We compute the no. of seconds and divide it to 60*60*24
    //We add one to inlude both dates in the interval.
    $days = ($endDate - $startDate) / 86400 + 1;

    $no_full_weeks = floor($days / 7);
    $no_remaining_days = fmod($days, 7);

    //It will return 1 if it's Monday,.. ,7 for Sunday
    $the_first_day_of_week = date("N", $startDate);
    $the_last_day_of_week = date("N", $endDate);

    //---->The two can be equal in leap years when february has 29 days, the equal sign is added here
    //In the first case the whole interval is within a week, in the second case the interval falls in two weeks.
    if ($the_first_day_of_week <= $the_last_day_of_week) {
        if ($the_first_day_of_week <= 6 && 6 <= $the_last_day_of_week) $no_remaining_days--;
        if ($the_first_day_of_week <= 7 && 7 <= $the_last_day_of_week) $no_remaining_days--;
    }
    else {
        // (edit by Tokes to fix an edge case where the start day was a Sunday
        // and the end day was NOT a Saturday)

        // the day of the week for start is later than the day of the week for end
        if ($the_first_day_of_week == 7) {
            // if the start date is a Sunday, then we definitely subtract 1 day
            $no_remaining_days--;

            if ($the_last_day_of_week == 6) {
                // if the end date is a Saturday, then we subtract another day
                $no_remaining_days--;
            }
        }
        else {
            // the start date was a Saturday (or earlier), and the end date was (Mon..Fri)
            // so we skip an entire weekend and subtract 2 days
            $no_remaining_days -= 2;
        }
    }

    //The no. of business days is: (number of weeks between the two dates) * (5 working days) + the remainder
//---->february in none leap years gave a remainder of 0 but still calculated weekends between first and last day, this is one way to fix it
   $workingDays = $no_full_weeks * 5;
    if ($no_remaining_days > 0 )
    {
      $workingDays += $no_remaining_days;
    }    


    return $workingDays;
}

// This will return 4
echo getWorkingDays("2012-09-06","2012-09-11");
?>

Answer:

If you don’t need full days but accurate seconds instead try this code. This accepts unix timestamps as an input.

function timeDifferenceWithoutWeekends($from, $to) {
    $start = new DateTime("@".$from);
    $current = clone $start;
    $end = new DateTime("@".$to);
    $sum = 0;
    while ($current<$end) {
        $endSlice = clone $current;
        $endSlice->setTime(0,0,0);
        $endSlice->modify('+1 day');
        if ($endSlice>$end) {
            $endSlice= clone $end;
        }
        $seconds = $endSlice->getTimestamp()-$current->getTimestamp();
        $currentDay = $current->format("D");
        if ($currentDay != 'Sat' && $currentDay != 'Sun') {
            $sum+=$seconds;
        }
        $current = $endSlice;
    }
    return $sum;
}

Answer:

/**
 * Getting the Weekdays count[ Excludes : Weekends]
 * 
 * @param type $fromDateTimestamp
 * @param type $toDateTimestamp
 * @return int
 */
public static function getWeekDaysCount($fromDateTimestamp = null, $toDateTimestamp=null) {

    $startDateString   = date('Y-m-d', $fromDateTimestamp);
    $timestampTomorrow = strtotime('+1 day', $toDateTimestamp);
    $endDateString     = date("Y-m-d", $timestampTomorrow);
    $objStartDate      = new \DateTime($startDateString);    //intialize start date
    $objEndDate        = new \DateTime($endDateString);    //initialize end date
    $interval          = new \DateInterval('P1D');    // set the interval as 1 day
    $dateRange         = new \DatePeriod($objStartDate, $interval, $objEndDate);

    $count = 0;

    foreach ($dateRange as $eachDate) {
        if (    $eachDate->format("w") != 6 
            &&  $eachDate->format("w") != 0 
        ) {
            ++$count;
        }
    }
    return $count;
}

Answer:

Here’s an alternative to calculate business days between two dates and also excludes USA holidays using Pear’s Date_Holidays from http://pear.php.net/package/Date_Holidays.

$start_date and $end_date should be DateTime objects (you can use new DateTime('@'.$timestamp) to convert from timestamp to DateTime object).

<?php
function business_days($start_date, $end_date)
{
  require_once 'Date/Holidays.php';
  $dholidays = &Date_Holidays::factory('USA');
  $days = 0;

  $period = new DatePeriod($start_date, new DateInterval('P1D'), $end_date);

  foreach($period as $dt)
  {
    $curr = $dt->format('D');

    if($curr != 'Sat' && $curr != 'Sun' && !$dholidays->isHoliday($dt->format('Y-m-d')))
    {
      $days++;
    }
  }
  return $days;
}
?>

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