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phpdatedatediff
Francisc
Francisc

Reputation: 80395

PHP get days between start-date and end-date

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

Thank you.

Upvotes: 5

Views: 16401

Answers (7)

Sandro Rosa
Sandro Rosa

Reputation: 517

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>" ;

?>

Upvotes: 1

lonesomeday
lonesomeday

Reputation: 237847

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

Upvotes: 6

ArK
ArK

Reputation: 21058

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

Upvotes: 3

ITroubs
ITroubs

Reputation: 11215

$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

Upvotes: 1

Etienne Marais
Etienne Marais

Reputation: 1690

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.

Upvotes: 1

Jichao
Jichao

Reputation: 41785

<?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);
?>

Upvotes: 2

Chinmayee G
Chinmayee G

Reputation: 8117

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";

Upvotes: 1

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