SparKot
0
Q:

php difference between two dates

$datetime1 	= new DateTime('2020-10-11 16:52:52');
$datetime2 	= new DateTime('2020-10-13 16:52:52');
$interval 	= $datetime1->diff($datetime2);

echo $interval->format('%a days');
1
$hourdiff = round((strtotime($time1) - strtotime($time2))/3600, 1);
2
//get Date diff as intervals 
$d1 = new DateTime("2018-01-10 00:00:00");
$d2 = new DateTime("2019-05-18 01:23:45");
$interval = $d1->diff($d2);
$diffInSeconds = $interval->s; //45
$diffInMinutes = $interval->i; //23
$diffInHours   = $interval->h; //8
$diffInDays    = $interval->d; //21
$diffInMonths  = $interval->m; //4
$diffInYears   = $interval->y; //1

//or get Date difference as total difference
$d1 = strtotime("2018-01-10 00:00:00");
$d2 = strtotime("2019-05-18 01:23:45");
$totalSecondsDiff = abs($d1-$d2); //42600225
$totalMinutesDiff = $totalSecondsDiff/60; //710003.75
$totalHoursDiff   = $totalSecondsDiff/60/60;//11833.39
$totalDaysDiff    = $totalSecondsDiff/60/60/24; //493.05
$totalMonthsDiff  = $totalSecondsDiff/60/60/24/30; //16.43
$totalYearsDiff   = $totalSecondsDiff/60/60/24/365; //1.35
6
$today = date("Y-m-d");
$expire = $row->expireDate; //from database

$today_time = strtotime($today);
$expire_time = strtotime($expire);

if ($expire_time < $today_time) { /* do Something */ }
0
$now = time(); // or your date as well
$your_date = strtotime("2010-01-31");
$datediff = $now - $your_date;

echo round($datediff / (60 * 60 * 24));
1
$period = new DatePeriod(
     new DateTime('2010-10-01'),
     new DateInterval('P1D'),
     new DateTime('2010-10-05')
);

//Which should get you an array with DateTime objects. 

//To iterate

foreach ($period as $key => $value) {
    //$value->format('Y-m-d')       
}
1
$date1 = "2007-03-24";
$date2 = "2009-06-26";

$diff = abs(strtotime($date2) - strtotime($date1));

$years = floor($diff / (365*60*60*24));
$months = floor(($diff - $years * 365*60*60*24) / (30*60*60*24));
$days = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24)/ (60*60*24));

printf("%d years, %d months, %d days\n", $years, $months, $days);
1

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