Answers for "time difference php"

PHP

datetime difference in php

//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

Posted by: Guest on November-01-2019

php get day diff

$now = time(); // or your date as well
$your_date = strtotime("2010-01-31");
$datediff = $now - $your_date;

echo round($datediff / (60 * 60 * 24));

Posted by: Guest on April-22-2020

Source

-1

calculate time difference php

$time1 = new DateTime('09:00:59');
$time2 = new DateTime('09:01:00');
$interval = $time1->diff($time2);
echo $interval->format('%s second(s)');

Result: 1 second(s)

Posted by: Guest on December-24-2020

difference of two dates in seconds php

$timeFirst  = strtotime('2011-05-12 18:20:20');
$timeSecond = strtotime('2011-05-13 18:20:20');
$differenceInSeconds = $timeSecond - $timeFirst;

Posted by: Guest on March-15-2021

Source

php time difference in hours

date_default_timezone_set("Africa/Johannesburg");
    $now = new DateTime();
    $future_date = new DateTime('2020-10-21 00:00:00');
    
    $interval = $future_date->diff($now);
    
    echo ($interval->format("%a") * 24) + $interval->format("%h"). " hours". $interval->format(" %i minutes ");
    print_r($now->format('Y-m-d H:i:s'));

Posted by: Guest on October-20-2020

Source

php difference between two dates

$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);

Posted by: Guest on July-18-2020

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