Calculating # of days in a date range that fall in a specific month

[anakadote]

Does anyone know a good way to calculate the number of days that fall in a specific month in a date range?

For example, in the date range 2012-04-14 - 2012-05-15, how many days are in the month of May?

I could just loop through each day in that range and increment a tally of how many days are in May, but is there a more efficient way of doing this?

$year = 2012;
$month = 5; // May
$date_start = strtotime('2012-04-14');
$date_end = strtotime('2012-05-15');

$num_days = 0;
while($date_start <= $date_end){
  if( date('Y', $date_start) == $year && date('n', $date_start) == $month) $num_days++;

  $date_start = strtotime('+1 day', $date_start);
}

echo $num_days . ' days';

Thanks.

[NogDog]

You could use http://php.net/manual/en/datetime.diff.php

[m_tt]

The number of days are in the date, why do any calculations?

echo date('d', strtotime('2012-05-15'))

[bradgrafelman]

m@tt;11015665 wrote:

The number of days are in the date, why do any calculations?

echo date('d', strtotime('2012-05-15'))

What if the given date range is 2012-04-14 - 2012-06-15?

[m_tt]

Well I think he would certainly need the loop in that case 🙂

$num_days = 0;
for($i = $date_start; $i <= $date_end; $i += 86400) {
	if (date('Y-m', $i) == '2012-05') {
		$num_days++;
	}
}

[m_tt]

You could also remove all functions from the loop. I'm also curious if there's an easier way to do this 😉

$match_year = 2012;
$match_month = 5;
$match_from = mktime(0, 0, 0, $match_month, 1, $match_year);
$match_to = mktime(0, 0, 0, $match_month, date('t', $match_from), $match_year);
$date_start = strtotime('2012-04-14');
$date_end = strtotime('2012-06-30');

$num_days = 0;
for($i = $date_start; $i <= $date_end; $i += 86400) {
	if ($i >= $match_from && $i <= $match_to) {
		$num_days++;
	}
}

[bradgrafelman]

m@tt;11015689 wrote:

$i += 86400

But what about the days which aren't 86,400 seconds in length?

[m_tt]

Are you saying there are days with only 23 hours? I'm not sure I understand the question.

60 60 24 = 86400

[bradgrafelman]

m@tt;11015703 wrote:

Are you saying there are days with only 23 hours?

Indeed; and others with 25 hours.

EDIT: And others still that are 24 hours and one second.

EDIT2: In fact, in that last case (days that are 24 hours and one second in length), the statement "1 minute == 60 seconds" isn't always true for that entire day.

[m_tt]

I have never heard of this, can you show me an example of a day that does not have 86400 seconds?

[bradgrafelman]

m@tt;11015707 wrote:

I have never heard of this, can you show me an example of a day that does not have 86400 seconds?

Sure! Let's go with this year (2012) for examples:

March 11th - DST begins - 23 hours in length
November 4th - DST ends - 25 hours in length
June 30th - Leap Second - 24 hours 1 second in length

EDIT: Forgot to mention - the first two assume you're in the United States.

[m_tt]

How would you calculate the actual seconds for those days? (Sorry OP to derail your thread 🙂)

echo mktime(0, 0, 0, 3, 12, 2012) - mktime(0, 0, 0, 3, 11, 2012); // 86400
echo mktime(0, 0, 0, 7, 1, 2012) - mktime(0, 0, 0, 6, 30, 2012); // 86400
echo mktime(0, 0, 0, 11, 5, 2012) - mktime(0, 0, 0, 11, 4, 2012); // 86400

[m_tt]

I have never heard of leap seconds before. I would still love to know how these can be problematic in PHP since I have dealt with timestamps a lot in the past. At least with the code I posted above, I can't cause any problems using 86400 seconds.

for($i = strtotime('2012-03-01'); $i <= strtotime('2012-04-01'); $i += 86400) { 
	echo date('Y-m-d H:i:s', $i) . PHP_EOL;
}

[bradgrafelman]

On my computer, this:

echo mktime(0, 0, 0, 11, 5, 2012) - mktime(0, 0, 0, 11, 4, 2012);

results in 90000 being output.

Your second test case is irrelevant since Unix timestamps don't observe leap seconds; instead, one "Unix second" actually lasts for two seconds in realtime.

EDIT: Feel free to use Google to research the effects of DST as well as leap seconds. I think we've de-railed the OP's thread enough at this point.

[NogDog]

And I'll still refer you back to my first reply, and let PHP do most of the work for you. 😉

<?php
$datetime1 = new DateTime('2012-04-14');
$datetime2 = new DateTime('2012-05-15');
$interval = $datetime1->diff($datetime2);
echo $interval->format('%a days');

[bradgrafelman]

@: That code doesn't answer the question posed in the OP's first post:

anakadote;11015659 wrote:

For example, in the date range 2012-04-14 - 2012-05-15, how many days are in the month of May?

[Derokorian]

Speaking of the OP's question. I just wrote this function to solve the question.

<?php

function DaysOfIntervalInMonth($IntervalStart,$IntervalEnd,$TestMonth,$TestYear = NULL)
{
   $IntervalStart = strtotime($IntervalStart);
   $IntervalEnd = strtotime($IntervalEnd);
   if( is_null($TestYear) && date('Y',$IntervalStart) == date('Y',$IntervalEnd) )
      $Year = date('Y',$IntervalStart);
   elseif ( !is_null($TestYear) )
      $Year = $TestYear;
   else
      return FALSE;

   if( !is_numeric($TestMonth) )
   {
      $TestMonth = date('M',strtotime("$TestMonth 01 2000"));
   }
   $MonthStart = $Year."-$TestMonth-01";
   $MonthStart = strtotime($MonthStart);
   $MonthEnd = strtotime("-1 Day",strtotime("+1 Month",$MonthStart));
   if( $MonthEnd < $IntervalStart || $MonthStart > $IntervalEnd )
      return 0;

   if( $IntervalStart > $MonthStart )
   {
      $MonthStart = $IntervalStart;
   }
   if( $IntervalEnd < $MonthEnd )
   {
      $MonthEnd = $IntervalEnd;
   }
   $Interval = date_diff(date_create("@$MonthStart"),date_create("@$MonthEnd"));
   return $Interval->d + 1;
}

var_dump(DaysOfIntervalInMonth('2012-03-21','2012-04-13','March'));
var_dump(DaysOfIntervalInMonth('2012-03-21','2012-04-13','April'));
var_dump(DaysOfIntervalInMonth('2012-03-21','2013-04-13','April',2012));
var_dump(DaysOfIntervalInMonth('2012-03-21','2012-04-13',4,2011));

Output:

php test.php
int(11)
int(13)
int(30)
int(0)

[NogDog]

<?php

function daysInLastMonthOfInterval($startDateStr, $endDateStr)
{
	$startDate = new DateTime($startDateStr);
	$endDate = new DateTime($endDateStr);
	$endMonthStart = new DateTime('first day of '.$endDateStr);
	$interval = $endDate->diff($endMonthStart);
	$range1 = ((int) $interval->format('%a')) + 1;
	$interval = $endDate->diff($startDate);
	$range2 = ((int) $interval->format('%a')) + 1;
	return min($range1, $range2);
}

$testCases = array(
	array(
		'2012-04-14',
		'2012-05-15'
	),
	array(
		'2012-05-10',
		'2012-05-15'
	),
	array(
		'2011-01-01',
		'2012-05-15'
	)
);
echo "<pre>";
foreach($testCases as $test)
{
	$result = daysInLastMonthOfInterval($test[0], $test[1]);
	echo $test[0].' => '.$test[1].' : '.$result."\n";
}
echo "</pre>";

2012-04-14 => 2012-05-15 : 15
2012-05-10 => 2012-05-15 : 6
2011-01-01 => 2012-05-15 : 15

[Derokorian]

Also for good measure I rewrote it using DateTime and related classes.

function DaysOfIntervalInMonth2($IntervalStart,$IntervalEnd,$TestMonth,$TestYear = NULL)
{
   $TimeZone = new DateTimeZone('UTC');
   $IntervalStart = new DateTime($IntervalStart,$TimeZone);
   $IntervalEnd = new DateTime($IntervalEnd,$TimeZone);
   if( is_null($TestYear) && $IntervalStart->format('Y') == $IntervalEnd->format('Y') )
      $TestYear = $IntervalStart->format('Y');
   elseif ( !is_null($TestYear) )
      $TestYear = $TestYear;
   else
      return FALSE;

   if( !is_numeric($TestMonth) )
   {
      $TestMonth = new DateTime("$TestMonth 01 2000",$TimeZone);
      $TestMonth = $TestMonth->format('M');
   }
   $MonthStart = new DateTime($TestYear."-$TestMonth-01",$TimeZone);
   $MonthEnd = new DateTime($TestYear."-$TestMonth-01",$TimeZone);
   $MonthEnd->add(new DateInterval('P1M'));
   $MonthEnd->sub(new DateInterval('P1D'));
   if( $MonthEnd < $IntervalStart || $MonthStart > $IntervalEnd )
      return 0;

   if( $IntervalStart > $MonthStart )
   {
      $MonthStart = $IntervalStart;
   }
   if( $IntervalEnd < $MonthEnd )
   {
      $MonthEnd = $IntervalEnd;
   }
   $Interval = $MonthStart->diff($MonthEnd);
   return $Interval->d + 1;
}

Edit: Also decided to benchmark the two versions, and the OO version is faster:

Running 2 Functions, with 6 sets of arguments, 1000 times each:
DaysOfIntervalInMonth
        Min: 0.000072
        Max: 0.001165
        Total: 0.810624
        Avg: 0.000135

DaysOfIntervalInMonth2
        Min: 0.000042
        Max: 0.000542
        Total: 0.378534
        Avg: 0.000063

[Weedpacket]

m@tt;11015713 wrote:

I have never heard of leap seconds before.

There have been 25 leap seconds added since 1972: every December 31 from 1972 to 1979; the December 31sts of 1987, 1989, 1990, 1995, 1998, 2005, and 2008; and the June 30ths of 1972, 1981, 1982, 1983, 1985, 1992, 1993, 1994, 1997, and 2012 have all been 86401 seconds long. So far there haven't been any 86399-second days, but they could happen should the Earth rotate faster than normal for some reason (the Earth's rotation is slowing down because of tidal braking, so the length of the year today is about a second longer than when the standard second was defined).

At least with the code I posted above, I can't cause any problems using 86400 seconds.

But you've heard of leap days, surely? Try

for($i = strtotime('2012-03-01'); $i <= strtotime('2013-03-01'); $i += 86400) { 
	echo date('Y-m-d H:i:s', $i) . PHP_EOL;
}

Of course, if you're configured to use a TZ that doesn't have a DST component, you won't see anything. Which is why:

1. Dates and times should be stored using UTC, and converted to local conventions only as and when needed, and

2. You shouldn't try to do your own date and time arithmetic if there's already a well-tested library available.