Subtracting Two Dates in PHP

How to calculate the difference between two dates using PHP?

Use this for legacy code (PHP < 5.3). For up to date solution see jurka's answer below

You can use strtotime() to convert two dates to unix time and then calculate the number of seconds between them. From this it's rather easy to calculate different time periods.

$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);

Edit: Obviously the preferred way of doing this is like described by jurka below. My code is generally only recommended if you don't have PHP 5.3 or better.

Several people in the comments have pointed out that the code above is only an approximation. I still believe that for most purposes that's fine, since the usage of a range is more to provide a sense of how much time has passed or remains rather than to provide precision - if you want to do that, just output the date.

Despite all that, I've decided to address the complaints. If you truly need an exact range but haven't got access to PHP 5.3, use the code below (it should work in PHP 4 as well). This is a direct port of the code that PHP uses internally to calculate ranges, with the exception that it doesn't take daylight savings time into account. That means that it's off by an hour at most, but except for that it should be correct.

<?php

/**
 * Calculate differences between two dates with precise semantics. Based on PHPs DateTime::diff()
 * implementation by Derick Rethans. Ported to PHP by Emil H, 2011-05-02. No rights reserved.
 * 
 * See here for original code:
 * http://svn.php.net/viewvc/php/php-src/trunk/ext/date/lib/tm2unixtime.c?revision=302890&view=markup
 * http://svn.php.net/viewvc/php/php-src/trunk/ext/date/lib/interval.c?revision=298973&view=markup
 */

function _date_range_limit($start, $end, $adj, $a, $b, $result)
{
    if ($result[$a] < $start) {
        $result[$b] -= intval(($start - $result[$a] - 1) / $adj) + 1;
        $result[$a] += $adj * intval(($start - $result[$a] - 1) / $adj + 1);
    }

    if ($result[$a] >= $end) {
        $result[$b] += intval($result[$a] / $adj);
        $result[$a] -= $adj * intval($result[$a] / $adj);
    }

    return $result;
}

function _date_range_limit_days($base, $result)
{
    $days_in_month_leap = array(31, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31);
    $days_in_month = array(31, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31);

    _date_range_limit(1, 13, 12, "m", "y", &$base);

    $year = $base["y"];
    $month = $base["m"];

    if (!$result["invert"]) {
        while ($result["d"] < 0) {
            $month--;
            if ($month < 1) {
                $month += 12;
                $year--;
            }

            $leapyear = $year % 400 == 0 || ($year % 100 != 0 && $year % 4 == 0);
            $days = $leapyear ? $days_in_month_leap[$month] : $days_in_month[$month];

            $result["d"] += $days;
            $result["m"]--;
        }
    } else {
        while ($result["d"] < 0) {
            $leapyear = $year % 400 == 0 || ($year % 100 != 0 && $year % 4 == 0);
            $days = $leapyear ? $days_in_month_leap[$month] : $days_in_month[$month];

            $result["d"] += $days;
            $result["m"]--;

            $month++;
            if ($month > 12) {
                $month -= 12;
                $year++;
            }
        }
    }

    return $result;
}

function _date_normalize($base, $result)
{
    $result = _date_range_limit(0, 60, 60, "s", "i", $result);
    $result = _date_range_limit(0, 60, 60, "i", "h", $result);
    $result = _date_range_limit(0, 24, 24, "h", "d", $result);
    $result = _date_range_limit(0, 12, 12, "m", "y", $result);

    $result = _date_range_limit_days(&$base, &$result);

    $result = _date_range_limit(0, 12, 12, "m", "y", $result);

    return $result;
}

/**
 * Accepts two unix timestamps.
 */
function _date_diff($one, $two)
{
    $invert = false;
    if ($one > $two) {
        list($one, $two) = array($two, $one);
        $invert = true;
    }

    $key = array("y", "m", "d", "h", "i", "s");
    $a = array_combine($key, array_map("intval", explode(" ", date("Y m d H i s", $one))));
    $b = array_combine($key, array_map("intval", explode(" ", date("Y m d H i s", $two))));

    $result = array();
    $result["y"] = $b["y"] - $a["y"];
    $result["m"] = $b["m"] - $a["m"];
    $result["d"] = $b["d"] - $a["d"];
    $result["h"] = $b["h"] - $a["h"];
    $result["i"] = $b["i"] - $a["i"];
    $result["s"] = $b["s"] - $a["s"];
    $result["invert"] = $invert ? 1 : 0;
    $result["days"] = intval(abs(($one - $two)/86400));

    if ($invert) {
        _date_normalize(&$a, &$result);
    } else {
        _date_normalize(&$b, &$result);
    }

    return $result;
}

$date = "1986-11-10 19:37:22";

print_r(_date_diff(strtotime($date), time()));
print_r(_date_diff(time(), strtotime($date)));

Subtracting two dates in php

Part 1: Why is the result 6?

The dates are simply strings when you first subtract them. PHP attempts to convert them to integers. It does this by converting until the first non-number. So, date2 become 6 and date1 becomes 0.

Part 2: How do you get it to work?

$datetime1 = strtotime('May 3, 2012 10:38:22 GMT');
$datetime2 = strtotime('06 Apr 2012 07:22:21 GMT');

$secs = $datetime2 - $datetime1;// == <seconds between the two times>
$days = $secs / 86400;

Convert as appropriate.

Calculate the difference between two dates in PHP?

A date difference by itself can only be precise to the Day. If you go up to the months and years, then the results will be inevitably false without a date reference.

You can say

9 years, 2 months, 29days till X date

and the reference would be today.

If you take the same period(Year, month, day), and use the date 2019-03-01 as reference, you would have the wrong result.

Here is an example:

$objDatetimeSource1 = new DateTime('2010-04-01');
$objDatetimeSource2 = new DateTime('2019-06-30');
$interval =  $objDatetimeSource1->diff($objDatetimeSource2);

$objDatetime1 = new DateTime('2020-02-28');
$objDatetime2 = new DateTime('2020-03-01');

$objDatetime1->add($interval); will give 2029-05-27 00:00:00 and

$objDatetime1->add($interval); will give 2029-05-30 00:00:00 with 3 days difference between them, when, from the dates 2020-02-28 and 2020-03-01, you can clearly see that there is only 2 days difference.

Php subtract two dates

Dates don't subtract like that. Convert both dates to time, subtract, and convert back to a date.

$now = strtotime(date("Y-m-d H:i:s"));
$status_date = strtotime($row['status_date']);
$x = date($now-$status_date);

However if you're trying to get the time between the dates consider the moment library.

https://github.com/fightbulc/moment.php

How to subtract two dates without sunday in php

You can find total days, and then find number of sundays in that range. Finally reduce those days from total days:

<?php    
$receipt_date   = new DateTime('2016-05-19');
$date           = new DateTime('2016-05-23');
$difference     = $receipt_date->diff($date);

$daterange = new DatePeriod($receipt_date, new DateInterval('P1D'), $date);

$count = 0;

foreach($daterange as $date){
    if(!($date->format("w"))) {
        $count++;
    }
}

echo ($difference->days)-$count;

Demo

Finding the number of days between two dates

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

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

How to subtract two dates and get the difference in minutes PHP

The easiest solution is probably to use strtotime and divide by 60:

$time = strtotime('01:47');
$time2 = strtotime('01:50');

$finalresult = ($time2 - $time) / 60;

Oracle : how to subtract two dates and get seconds of the result

Your data type is most probably TIMESTAMP that would explain the rounding problem.

You may workaround it by first casting to DATE (to get rid of the milliseconds) and that casting it back to TIMESTAMP (to be able to perform your regexp_substr)

This sample data replays your problem

select opa.*, 
NVL(REGEXP_SUBSTR (CAST(opa.end_time AS TIMESTAMP) - CAST(opa.start_time AS TIMESTAMP), '\d{2}:\d{2}:\d{2}'),' ') AS duration,
NVL(REGEXP_SUBSTR (CAST(CAST(opa.end_time AS DATE)AS TIMESTAMP) - CAST(CAST(opa.start_time AS DATE)AS TIMESTAMP), '\d{2}:\d{2}:\d{2}'),' ') AS duration2
from tab opa;

START_TIME                           END_TIME                             DURATION                    DURATION2                   
------------------------------------ ------------------------------------ --------------------------- ---------------------------
04.05.2021 09:13:07,555000000 +02:00 04.05.2021 09:13:18,111000000 +02:00 00:00:10                    00:00:11 

PHP How to subtract two dates with milliseconds

Here is a solution 5.2.2 < php version < 7.1

<?php
$date1=\DateTime::createFromFormat('Y-m-d H:i:s.u','2017-12-13 13:06:21.602');
$date2=\DateTime::createFromFormat('Y-m-d H:i:s.u','2017-12-13 13:06:18.102');
// $di =  $date1->diff($date2);
// var_dump( $di );

// seconds
$diffSeconds = $date1->getTimestamp() - $date2->getTimestamp();
// milli
$diffMilli = ($date1->format('u') - $date2->format('u')) / 1000000;
//result :
$diffWithMilli = $diffSeconds + $diffMilli;
var_dump($diffWithMilli);

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