DateInterval total number of days as float

Question

For now I don't care about timezones and just want to have a correct calculated value of how many days are between two dates. I think I'm pretty near to what I want, but for me there is always a full day missing.

That's what I got so far:

$start = new DateTimeImmutable('2018-07-31 12:30:00');
$end = new DateTimeImmutable('2018-08-28 00:00:00');
$diff = $start->diff($end);

$totalDays = (int) $diff->days; // 27
$hours = $diff->h;              // 11
$minutes = $diff->i;            // 30
$seconds = $diff->s;            //  0

$hoursAsDays = $hours / 24;               // 0.45833333333333
$minutesAsDays = $minutes / 60 / 24;      // 0.020833333333333
$secondsAsDays = $seconds / 60 / 60 / 24; // 0
$result = $totalDays + $hoursAsDays + $minutesAsDays + $secondsAsDays;

var_dump($result); // 27.479166666667

For me there are 28 full days from august. I thought I would get something like: 28.45

EDIT: Users who add '2018-08-28' as end date will assume that it will count as a full day. How do I safely make this possible to be mentioned in my calculation?

the code is correctly working:- difference is 27 days 11 hours 30 min 0 second. Conversion can modified to look correct:- https://eval.in/939764 — Alive to die - Anant, Jan 22 '18 at 08:43
there are not full 28 days of august `31 12:30:00` will treated as half day+30 minutes of 1 ugust — Alive to die - Anant, Jan 22 '18 at 08:45
@AlivetoDie, No its not including the half day of 1st august, it is just taking 27 days of august (because OP is taking just the start of the 28th august) — Shubhranshu, Jan 22 '18 at 08:47
@AlivetoDie I agree with the answer but not with the fact , 31 12:30:00 will treated as half day+30 minutes of 1 august — Shubhranshu, Jan 22 '18 at 08:51

user2342558 · Accepted Answer · 2023-03-08T08:20:09.307

You can do that simply with strtotime() PHP function:

$start = '2018-07-31 12:30:00';
$end = '2018-08-28 00:00:01';

echo $diffDays = (strtotime($end) - strtotime($start)) / 60 / 60 / 24;

Result:

27.479166666667

The result is true because there are exactly 27,5 days (in terms of 24 hours per day) between the two dateTime:

27 days and early a half day:
from 2018-07-31 12:30 to 2018-07-31 23.59 = +1/2 day tot:    0,5
from 2018-08-01 00:00 to 2018-08-01 23.59 = +1 day   tot:    1,5
from 2018-08-02 00:00 to 2018-08-01 23.59 = +1 day   tot:    2,5
from 2018-08-03 00:00 to 2018-08-01 23.59 = +1 day   tot:    3,5
from 2018-08-04 00:00 to 2018-08-01 23.59 = +1 day   tot:    4,5
from 2018-08-05 00:00 to 2018-08-01 23.59 = +1 day   tot:    5,5
from 2018-08-06 00:00 to 2018-08-01 23.59 = +1 day   tot:    6,5
from 2018-08-07 00:00 to 2018-08-01 23.59 = +1 day   tot:    7,5
from 2018-08-08 00:00 to 2018-08-01 23.59 = +1 day   tot:    8,5
from 2018-08-09 00:00 to 2018-08-01 23.59 = +1 day   tot:    9,5
from 2018-08-10 00:00 to 2018-08-01 23.59 = +1 day   tot:   10,5
from 2018-08-11 00:00 to 2018-08-01 23.59 = +1 day   tot:   11,5
from 2018-08-12 00:00 to 2018-08-01 23.59 = +1 day   tot:   12,5
from 2018-08-13 00:00 to 2018-08-01 23.59 = +1 day   tot:   13,5
from 2018-08-14 00:00 to 2018-08-01 23.59 = +1 day   tot:   14,5
from 2018-08-15 00:00 to 2018-08-01 23.59 = +1 day   tot:   15,5
from 2018-08-16 00:00 to 2018-08-01 23.59 = +1 day   tot:   16,5
from 2018-08-17 00:00 to 2018-08-01 23.59 = +1 day   tot:   17,5
from 2018-08-18 00:00 to 2018-08-01 23.59 = +1 day   tot:   18,5
from 2018-08-19 00:00 to 2018-08-01 23.59 = +1 day   tot:   19,5
from 2018-08-20 00:00 to 2018-08-01 23.59 = +1 day   tot:   20,5
from 2018-08-21 00:00 to 2018-08-01 23.59 = +1 day   tot:   21,5
from 2018-08-22 00:00 to 2018-08-01 23.59 = +1 day   tot:   22,5
from 2018-08-23 00:00 to 2018-08-01 23.59 = +1 day   tot:   23,5
from 2018-08-24 00:00 to 2018-08-01 23.59 = +1 day   tot:   24,5
from 2018-08-25 00:00 to 2018-08-01 23.59 = +1 day   tot:   25,5
from 2018-08-26 00:00 to 2018-08-01 23.59 = +1 day   tot:   26,5
from 2018-08-27 00:00 to 2018-08-01 23.59 = +1 day   tot:   27,5
from 2018-08-28 00:00 to 2018-08-28 00:00 = +0 day   tot:   27,5

With strtotime() you can convert a date/datetime string into instant version (count from epoch in seconds).

Regarding your edit, to consider the ending day in the calculation, set its time as 23:59:59:

$end = date('Y-m-d 23:59:59', strtotime('2018-08-28 00:00:01'));

Thus, the result will be:

28.479155092593

That makes sense. Thank you for the explanation. But users to add '2018-08-28' as end date will asume that it will count as a full day. How do I safely make this possible to be mentioned in my calculation? I'll add this question to the initial post. — simplychrislike, Jan 22 '18 at 09:34

score 3 · Answer 2 · answered Jan 22 '18 at 10:56

Just to have this complete, this is my result now:

/**
 * @param DateTimeInterface $start
 * @param DateTimeInterface $end
 * @return float|int
 */
public static function getDaysBetweenDates(DateTimeInterface $start, DateTimeInterface $end)
{
    $start = new DateTimeImmutable("@{$start->getTimestamp()}");
    $end = new DateTimeImmutable("@{$end->getTimestamp()}");
    $diff = $start->diff($end);

    if (
        (int) $end->format('H') === 0
        && (int) $end->format('i') === 0
        && (int) $end->format('s') === 0
    ) {
        $end = new DateTimeImmutable($end->format('Y-m-d') . ' 23:59:59');
        $diff = $start->diff($end);
    }

    $totalDays = (int) $diff->days;
    $hours = $diff->h;
    $minutes = $diff->i;
    $seconds = $diff->s;

    $hoursAsDays = $hours / 24;
    $minutesAsDays = $minutes / 60 / 24;
    $secondsAsDays = $seconds / 60 / 60 / 24;
    return $totalDays + $hoursAsDays + $minutesAsDays + $secondsAsDays;
}

score 0 · Answer 3 · answered May 04 '20 at 15:53

0

Check the %a option in format method of DateInterval

$date1 = new DateTime('19-04-2019');
$date2 = new DateTime('5-05-2020');
echo $date2->diff($date1)->format('%a days');

output:

382 days

answered May 04 '20 at 15:53

Grigory Ilizirov

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DateInterval total number of days as float

3 Answers3