how to find number of mondays or tuesdays between two dates?

0 votes
asked Oct 31, 2009 by svk

I have start date and end date.

I need to find out the day that is Sunday or Monday etc dependent upon user click on check box.

How can I find/calculate that in PHP?

8 Answers

0 votes
answered Oct 31, 2009 by svk

I got the answer.Its working for sunday only.But I dont know how to make it for another days

"; $days_until_sunday = date('w', $start) > 0 ? 7 - date('w', $start) : 0; $date = $start + (ONE_DAY * $days_until_sunday); $sundays = array(); while ($date "; $count=0; // Loop and output Y-m-d foreach (sundays_in_range($start, $end) as $sunday) { print "".date("Y-m-d", $sunday)."
"; $count++; } echo $count; ?>
0 votes
answered Oct 31, 2009 by w35l3y

no loops and no recursivity

<?php
define('ONE_WEEK', 604800); // 7 * 24 * 60 * 60

function number_of_days($days, $start, $end) {
    $w = array(date('w', $start), date('w', $end));
    $x = floor(($end-$start)/ONE_WEEK);
    $sum = 0;

    for ($day = 0;$day < 7;++$day) {
        if ($days & pow(2, $day)) {
            $sum += $x + ($w[0] > $w[1]?$w[0] <= $day || $day <= $w[1] : $w[0] <= $day && $day <= $w[1]);
        }
    }

    return $sum;
}

//$start = $end = time();

// 0x10 == pow(2, 4) == 1 << 4 // THURSDAY
// 0x20 == pow(2, 5) == 1 << 5 // FRIDAY
echo number_of_days(0x01, $start, $end); // SUNDAY
echo number_of_days(pow(2, 0), $start, $end); // SUNDAY
echo number_of_days(0x02, $start, $end); // MONDAY
echo number_of_days(pow(2, 1), $start, $end); // MONDAY
echo number_of_days(0x04, $start, $end); // TUESDAY
echo number_of_days(1 << 2, $start, $end); // TUESDAY
echo number_of_days(0x08, $start, $end); // WEDNESDAY
echo number_of_days(1 << 3, $start, $end); // WEDNESDAY
echo number_of_days(0x10, $start, $end); // THURSDAY
echo number_of_days(0x20, $start, $end); // FRIDAY
echo number_of_days(0x40, $start, $end); // SATURDAY
echo number_of_days(0x01 | 0x40, $start, $end); // WEEKENDS : SUNDAY | SATURDAY
echo number_of_days(0x3E, $start, $end); // WORKDAYS : MONDAY | TUESDAY | WEDNESDAY | THURSDAY | FRIDAY
?>
0 votes
answered Oct 28, 2010 by phil-mccull

The answer by w35I3y was almost correct, but I was getting errors using that function. This function correctly calculates the number of Mondays or any specific day between two given dates:

/** 
* Counts the number occurrences of a certain day of the week between a start and end date
* The $start and $end variables must be in UTC format or you will get the wrong number 
* of days  when crossing daylight savings time
* @param - $day - the day of the week such as "Monday", "Tuesday"...
* @param - $start - a UTC timestamp representing the start date
* @param - $end - a UTC timestamp representing the end date
* @return Number of occurences of $day between $start and $end
*/
function countDays($day, $start, $end)
{        
    //get the day of the week for start and end dates (0-6)
    $w = array(date('w', $start), date('w', $end));

    //get partial week day count
    if ($w[0] < $w[1])
    {            
        $partialWeekCount = ($day >= $w[0] && $day <= $w[1]);
    }else if ($w[0] == $w[1])
    {
        $partialWeekCount = $w[0] == $day;
    }else
    {
        $partialWeekCount = ($day >= $w[0] || $day <= $w[1]);
    }

    //first count the number of complete weeks, then add 1 if $day falls in a partial week.
    return floor( ( $end-$start )/60/60/24/7) + $partialWeekCount;
}

Example Usage:

$start = strtotime("tuesday UTC");    
$end = strtotime("3 tuesday UTC");       
echo date("m/d/Y", $start). " - ".date("m/d/Y", $end). " has ". countDays(0, $start, $end). " Sundays";

Outputs something like: 09/28/2010 - 10/19/2010 has 3 Sundays.

0 votes
answered Oct 8, 2011 by gaurav-srivastava
<?php
$date = strtotime('2009-01-01');
$dateMax = strtotime('2009-02-23');

$nbr = 0;
while ($date < $dateMax) {
  var_dump(date('Y-m-d', $date));
  $nbr++;
  $date += 7 * 24 * 3600;
}
echo "<pre>";
 var_dump($nbr);
?>
0 votes
answered Oct 10, 2012 by dvyper
function daycount($day, $startdate, $enddate, $counter) {
    if($startdate >= $enddate) {
        return $counter-1;  // A hack to make this function return the correct number of days.
    } else {
        return $this->daycount($day, strtotime("next ".$day, $startdate), $enddate, ++$counter);
    }
}

This is a different version of the first answer which takes a start and end point and works for me. All of the examples given on this page seemed to return the answer plus an additional day for some reason.

0 votes
answered Oct 2, 2013 by overflowing

w35l3y's answer seems to work well, so I up-voted it. Still, I preferred something a bit easier to follow, and with less math and looping (not sure it matters from a performance standpoint, though). I think I covered all the possible scenarios... Here's what I came up with:

function numDays($sday, $eday, $i, $cnt) {
    if (($sday < $eday && $i >= $sday && $i <= $eday) || ($sday > $eday && ($i >= $sday || $i <= $eday))) {
        // partial week (implied by $sday != $eday), so $i (day iteration) may have occurred one more time
        // a) end day is ahead of start day; $i is within start/end of week range
        // b) start day is ahead of end day (i.e., Tue start, Sun end); $i is either in back half of first week or front half of second week
        $cnt++;
    } elseif ($sday == $eday && $i == $sday) {
        // start day and end day are the same, and $i is that day, i.e., Tue occurs twice from Tue-Tue (1 wk, so $wks = $cnt)
        $cnt++;
    }

    return $cnt;    // # of complete weeks + partial week, if applicable
}

Notes: $sday and $eday are the day numbers corresponding to the start and end of the range to be checked, and $i is the day number being counted (I have it in a 0-6 loop). I moved $wks outside the function, as there's no point recalculating it each time.

$wks = floor(($endstamp - $startstamp)/7*24*60*60);
$numDays = numDays($sday, $eday, $i, $wks);

Make sure the start/end timestamps you're comparing have the same timezone adjustment, if any, otherwise you'll always be off a bit with $cnt and $wks. (I ran into that when counting from an unadjusted first of the year to an adjusted day/time X.)

0 votes
answered Oct 8, 2013 by vascowhite

This question is just crying out for an updated answer that uses PHP's DateTime classes, so here it is:-

/**
 * @param String $dayName eg 'Mon', 'Tue' etc
 * @param DateTimeInterface $start
 * @param DateTimeInterface $end
 * @return int
 */
function countDaysByName($dayName, \DateTimeInterface $start, \DateTimeInterface $end)
{
    $count = 0;
    $interval = new \DateInterval('P1D');
    $period = new \DatePeriod($start, $interval, $end);

    foreach($period as $day){
        if($day->format('D') === ucfirst(substr($dayName, 0, 3))){
            $count ++;
        }
    }
    return $count;
}
0 votes
answered Oct 30, 2014 by lakhan-gehlot

To count The Total Friday Between Two Dates##

$from_date=(2015-01-01);

$to_date=(2015-01-20);

while(strtotime($from_date) <= strtotime($to_date)){

    //5 for count Friday, 6 for Saturday , 7 for Sunday

    if(date("N",strtotime($from_date))==5){
        $counter++;
    }
    $from_date = date ("Y-m-d", strtotime("+1 day", strtotime($from_date)));

}

echo $counter;
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