PHP get days between start-date and end-date

Question

If I have two variables $startDate="YYYYmmdd" and $endDate="YYYYmmdd", how can I get the number of days between them please?

Thank you.

please indicate if you want number of days inclusive or exclusive of the two dates. — bcosca, Oct 13 '10 at 08:35

score 6 · Accepted Answer · edited Nov 13 '11 at 16:18

6

If you are using PHP 5.3, you can use the new DateTime class:

$startDate = new DateTime("20101013");
$endDate = new DateTime("20101225");

$interval = $startDate->diff($endDate);

echo $interval->days . " until Christmas"; // echos 73 days until Christmas

If not, you will need to use strtotime:

$startDate = strtotime("20101013");
$endDate = strtotime("20101225");

$interval = $endDate - $startDate;
$days = floor($interval / (60 * 60 * 24));

echo $days . " until Christmas"; // echos 73 days until Christmas

edited Nov 13 '11 at 16:18

Josh

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answered Oct 13 '10 at 08:18

lonesomeday

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The new DateTime, DateTimeInterval, DateTimePeriod and DateTimeZone-classes just rock. I also made some experience with them. I think it's better than strtotime, because it considers leap-years, leap-seconts, etc. So +1 for you. I also gave an more detailed example in this post: http://stackoverflow.com/questions/3108591/calculate-number-of-hours-between-2-dates-in-php/3108800#3108800 – Fidi Oct 13 '10 at 08:40

ArK · Answer 2 · 2010-10-13T08:24:37.087

3

<?php   
 $time1=strtotime($startDate);
    $time2=strtotime($endDate);
    $daycount=floor(($time2-$time1)/ 86400);
?>

edited Oct 13 '10 at 08:24

answered Oct 13 '10 at 08:19

ArK

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score 2 · Answer 3 · answered Oct 13 '10 at 08:19

2

<?php
function days($date1, $date2) {
    $date1 = strtotime($date1);
    $date2 = strtotime($date2);
    return ($date2 - $date1) / (24 * 60 * 60);
}
$date1 = '20100820';
$date2 = '20100930';
echo days($date1, $date2);
?>

answered Oct 13 '10 at 08:19

Jichao

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score 1 · Answer 4 · answered Sep 06 '15 at 20:34

Here's my approach, based upon a brutal search in most cases, just because divisions by seconds (for weeks, months, years) may not return precise results, as while working with leap years for example.

<?php
function datediff( $timeformat, $startdate, $enddate )
{
    $unix_startdate = strtotime( $startdate ) ;
    $unix_enddate = strtotime( $enddate ) ;
    $min_date = min($unix_startdate, $unix_enddate);
    $max_date = max($unix_startdate, $unix_enddate);
    $Sd = date( "d", $unix_startdate ) ;
    $Sm = date( "m", $unix_startdate ) ;
    $Sy = date( "Y", $unix_startdate ) ;
    $Ed = date( "d", $unix_enddate ) ;
    $Em = date( "m", $unix_enddate ) ;
    $Ey = date( "Y", $unix_enddate ) ;

    $unixtimediff = $unix_enddate - $unix_startdate ;
    if ( $unixtimediff <= 0 ) return -1 ;

    switch( strtolower( $timeformat ) )
    {
         case "d": // days
         $divisor = 3600 * 24 ;
         return floor( $unixtimediff / $divisor ) + 1 ; 
         break ;
         case "w": // weeks
         $i = 0 ;
         while ( ( $min_date = strtotime("+1 DAY", $min_date) ) <= $max_date) $i++;
         return floor( $i / 7 ) ;
         break ;
         case "m": // months
         $i = $Sd != $Ed && $Sm != $Em ? 1 : 0 ;
         while ( ( $min_date = strtotime("+1 MONTH", $min_date) ) <= $max_date) $i++;
         return $i ;
         break ;
         case "q": // quaterly (3 months)
         $i = $Sd != $Ed && $Sm != $Em ? 1 : 0 ;
         while ( ( $min_date = strtotime("+3 MONTH", $min_date) ) <= $max_date) $i++;
         return $i ;
         break ;
         case "y": // year
         $i = $Sd != $Ed && $Sm != $Em ? 1 : 0 ;
         while ( ( $min_date = strtotime("+1 MONTH", $min_date) ) <= $max_date) $i++;
         return floor( $i / 12 ) ;
         break ;
    }
}

$startdate = "2014-01-01" ;
$enddate = "2015-12-31" ;
$formats = array( "d" => "days", "w" => "weeks", "m" => "months", "q" => "quaterly", "y" => "years" ) ;
foreach( $formats AS $K => $F )
echo "From $startdate to $enddate in $F format: ". datediff( "$K",  $startdate, $enddate )."<br>" ;

?>

ITroubs · Answer 5 · 2010-10-13T08:24:01.377

1

$DayDiff = strtotime("2010-01-12")-strtotime("2009-12-30");
echo  date('z', $DayDiff)." Days";

this one should be precise and usable with PHP < 5.2

edited Oct 13 '10 at 08:24

answered Oct 13 '10 at 08:17

ITroubs

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score 1 · Answer 6 · answered Oct 13 '10 at 08:17

1

Here is the sample code

$startDate = mktime(0,0,0,1,1,2010); 
$endDate = mktime(0,0,0,12,1,2010); 

$dateDiff = $date1 - $date2;
$fullDays = floor($dateDiff/(60*60*24));
echo "Differernce is $fullDays days";

answered Oct 13 '10 at 08:17

Chinmayee G

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score 1 · Answer 7 · answered Oct 13 '10 at 08:19

1

The easiest way I have found to get the number of days between them is by converting the Start and End dates to Unix timestamps and doing an subtract on them.

Then if you want to format the date convert it back using the PHP date function.

answered Oct 13 '10 at 08:19

Etienne Marais

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PHP get days between start-date and end-date

7 Answers7