What can use for DateTime::diff() for PHP 5.2?

Question

Is there any function equivalent to DateTime::diff() in PHP 5.2?

My local server is PHP 5.3 and using DateTime::diff(). then I found that my live site uses PHP 5.2 and gives an error.

Fatal error: Call to undefined method DateTime::diff() in /var/www/some/other/dir/web/daikon/modules/projects/views/admin/log/admin_log_list.php on line 40

The PHP code:

 foreach ($logs as $key => $list){
 ...
 // show date in European way dd-mm-yyyy not in MySQL way yyyy-mm-dd
    $newdate =new DateTime($list['date']) ;
    echo "<td class=\"left\" width=\"8%\">".$newdate->format('d-m-Y')."</td>\n";
    $starttime = new DateTime($list['start_time']);
    echo "<td width=\"7%\">".date_format($starttime, 'H:i')."</td>\n";
    $finishtime = new DateTime($list['finish_time']);
    echo "<td width=\"8%\">".date_format($finishtime, 'H:i')."</td>\n";
    $timediff = 0;
    $interval = $starttime->diff($finishtime);
    $hours   = $interval->format('%h');
    $minutes = $interval->format('%i');
    $timediff = $hours * 60 + $minutes;

score 41 · Answer 1 · edited Jun 20 '13 at 20:24

41

Spudley's answer doesn't work for me--subtracting any DateTime from another gives 0 on my system.

I was able to get it to work by using DateTime::format with the 'U' specifier (seconds since Unix epoch):

$start = new DateTime('2010-10-12');
$end = new DateTime();
$days = round(($end->format('U') - $start->format('U')) / (60*60*24));

This works on both my dev system (5.3.4) and my deployment system (5.2.11).

edited Jun 20 '13 at 20:24

senia

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answered Apr 05 '11 at 06:39

Christopher Pickslay

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2

Almost perfect, except you used `abs()`, which assume it is always positive. – Raptor Mar 14 '13 at 03:29
1

Thanks for pointing that out, @ShivanRaptor. I looked back at the code I wrote this for, and it requires a positive value, and always compares today with a date in the past. The bug in that code was that I subtracted today from the start date, instead of the other way around, so I used `abs()` to convert it to a positive value. I've updated the sample above to return a negative value if the end date is before the start date, which is the default behavior of `DateTime::diff()`. – Christopher Pickslay Mar 14 '13 at 18:29
I needed to add perenthesis to the upper part of division: `$days = round(($end->format('U') - $start->format('U')) / (60*60*24))` – K. Weber May 08 '13 at 21:30
This not working properly, using DateTime::diff() I got 340 days and with this code 341. Don't recommend to use it on production. – Oleg Sapishchuk Nov 02 '15 at 11:45

score 6 · Answer 2 · answered Jul 09 '12 at 06:17

I just needed that ( unfortunately ) for a WordPress plugin. This I use the function in 2 times:

In my class calling ->diff() ( my class extends DateTime, so $this is the reference DateTime )

function diff ($secondDate){
    $firstDateTimeStamp = $this->format("U");
    $secondDateTimeStamp = $secondDate->format("U");
    $rv = ($secondDateTimeStamp - $firstDateTimeStamp);
    $di = new DateInterval($rv);
    return $di;
}

Then I recreated a fake DateInterval class ( because DateInterval is only valid in PHP >= 5.3 ) as follows:

Class DateInterval {
    /* Properties */
    public $y = 0;
    public $m = 0;
    public $d = 0;
    public $h = 0;
    public $i = 0;
    public $s = 0;

    /* Methods */
    public function __construct ( $time_to_convert /** in seconds */) {
        $FULL_YEAR = 60*60*24*365.25;
        $FULL_MONTH = 60*60*24*(365.25/12);
        $FULL_DAY = 60*60*24;
        $FULL_HOUR = 60*60;
        $FULL_MINUTE = 60;
        $FULL_SECOND = 1;

//        $time_to_convert = 176559;
        $seconds = 0;
        $minutes = 0;
        $hours = 0;
        $days = 0;
        $months = 0;
        $years = 0;

        while($time_to_convert >= $FULL_YEAR) {
            $years ++;
            $time_to_convert = $time_to_convert - $FULL_YEAR;
        }

        while($time_to_convert >= $FULL_MONTH) {
            $months ++;
            $time_to_convert = $time_to_convert - $FULL_MONTH;
        }

        while($time_to_convert >= $FULL_DAY) {
            $days ++;
            $time_to_convert = $time_to_convert - $FULL_DAY;
        }

        while($time_to_convert >= $FULL_HOUR) {
            $hours++;
            $time_to_convert = $time_to_convert - $FULL_HOUR;
        }

        while($time_to_convert >= $FULL_MINUTE) {
            $minutes++;
            $time_to_convert = $time_to_convert - $FULL_MINUTE;
        }

        $seconds = $time_to_convert; // remaining seconds
        $this->y = $years;
        $this->m = $months;
        $this->d = $days;
        $this->h = $hours;
        $this->i = $minutes;
        $this->s = $seconds;
    }
}

Hope that helps somebody.

nice, but why not use the mod function (`%`) (and integer division) instead of looping to get remainder — Nikos M., Jul 21 '15 at 18:16

score 3 · Answer 3 · edited Oct 20 '12 at 12:03

I was trying to improve Christopher Pickslay's answer.

I made this function that returns and object with most of the properties from the original DateInterval object.

There is no "days" property, because it seems to be some bug in my test server (it always return 6015) .

Also, I am assuming every month has 30 days, which is definitely not precise, but may help.

function dateTimeDiff($date1, $date2) {

    $alt_diff = new stdClass();
    $alt_diff->y =  floor(abs($date1->format('U') - $date2->format('U')) / (60*60*24*365));
    $alt_diff->m =  floor((floor(abs($date1->format('U') - $date2->format('U')) / (60*60*24)) - ($alt_diff->y * 365))/30);
    $alt_diff->d =  floor(floor(abs($date1->format('U') - $date2->format('U')) / (60*60*24)) - ($alt_diff->y * 365) - ($alt_diff->m * 30));
    $alt_diff->h =  floor( floor(abs($date1->format('U') - $date2->format('U')) / (60*60)) - ($alt_diff->y * 365*24) - ($alt_diff->m * 30 * 24 )  - ($alt_diff->d * 24) );
    $alt_diff->i = floor( floor(abs($date1->format('U') - $date2->format('U')) / (60)) - ($alt_diff->y * 365*24*60) - ($alt_diff->m * 30 * 24 *60)  - ($alt_diff->d * 24 * 60) -  ($alt_diff->h * 60) );
    $alt_diff->s =  floor( floor(abs($date1->format('U') - $date2->format('U'))) - ($alt_diff->y * 365*24*60*60) - ($alt_diff->m * 30 * 24 *60*60)  - ($alt_diff->d * 24 * 60*60) -  ($alt_diff->h * 60*60) -  ($alt_diff->i * 60) );
    $alt_diff->invert =  (($date1->format('U') - $date2->format('U')) > 0)? 0 : 1 ;

    return $alt_diff;
}

with this instead 30 we can calculate the exact days the month has: cal_days_in_month(CAL_GREGORIAN, 8, 2003); — Alejandro Moreno, Aug 27 '13 at 10:31

score 3 · Answer 4 · answered Mar 02 '12 at 12:31

I use this, seems to work alright - obviously you can add a second parameter to make it more flexible:

function GetDateDiffFromNow($originalDate) 
{
    $unixOriginalDate = strtotime($originalDate);
    $unixNowDate = strtotime('now');
    $difference = $unixNowDate - $unixOriginalDate ;
    $days = (int)($difference / 86400);
    $hours = (int)($difference / 3600);
    $minutes = (int)($difference / 60);
    $seconds = $difference;

    // now do what you want with this now and return ...
}

score 2 · Answer 5 · answered Oct 27 '10 at 12:47

Yes, it's annoying that feature didn't make it into PHP5.2.

I'll assume you can't upgrade to 5.3? You should look into it; there's very little reason not to upgrade; but I'll assume you can't for whatever the reason.

First tip: If you only need a diff of less than 24hours, you can simply subtract the two time stamps, and do $time_diff = date('H:i:s',$subtracted_value);

If you're doing more than 24 hour diffs, but you're okay with just returning the number of days along with the time difference, you can expand on the above technique by doing a modulus calculation on the subtrated value, against the number of seconds in a day (ie 24*60*60, which is 86400)

$subtracted_value = $date1 - $date2;
$days_diff = $subtracted_value % 86400;
$time_diff = date('H:i:s',$subtracted_value);

If you need weeks, you can of course do $days_diff % 7.

Unfortunately, the manual technique breaks down after weeks, because months and years are variable in length (technically days are too, given daylight saving, but you can probably ignore that, especially as you're only ever going to be one hour out, at the most), but hopefully that's good enough to get you started.

@Rocket.... hmm, possibly. This answer was a while ago, so I can't remember what I was thinking, but you may well be right. — Spudley, Jul 13 '11 at 07:49
I realize this was an old answer, but I found it on Google after searching for how to use `DateTime::diff()` on PHP 5.2. — gen_Eric, Jul 13 '11 at 13:11

score 2 · Answer 6 · answered Feb 17 '11 at 21:25

2

Read the contrib notes in the PHP date_diff page http://us3.php.net/manual/en/function.date-diff.php

answered Feb 17 '11 at 21:25

ZCJ

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score 2 · Answer 7 · answered Jul 04 '11 at 00:02

2

Spudley was close, but you need to use gmdate not date.

So this works for 24 hours or less (if it's a positive value at least):

$difference = $date2->format('U') - $date1->format('U');
$time_diff = gmdate('H:i:s',$difference);

answered Jul 04 '11 at 00:02

Mark

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score 0 · Answer 8 · answered Sep 26 '12 at 10:38

Please observe that if your DateTime object was created from a date string without any timezone information (as in '2012-01-01 05:00:00' like from mysql), then setting the timezone later with DateTimeZone objects via setTimeZone() does not change the DateTime objects internal timestamp.

$localtime = new DateTime('2012-01-01 08:00:00+02:00'); // Europe/Copenhagen (+ daylight savings)
$localtime->setTimezone(new DateTimeZone('UTC')); // convert local time to utc

$utctime = new DateTime('2012-01-01 06:00:00'); // timezone assumed utc, but is in fact unknown
$utctime->setTimezone(new DateTimeZone('UTC')); // trying to rectify missing timezone which fails, because the internal timestamp isn't modified, although any other format than 'U' may indicate so
#$utctime = new DateTime('2012-01-01 06:00:00+00:00'); // timezone stated, this works

$diff = intval($localtime->format('U')) - intval($utctime->format('U'));
echo $diff; // expecting zero

score 0 · Answer 9 · edited Feb 24 '16 at 06:35

Here is the best solution for your question.

<?php

/* Find difference between two dates in days. PHP 5.2.x.  */

$d1 = date('Y-m-d', strtotime('2013-06-13'));
$d2 = date("Y-m-d");
echo diff($d1, $d2);

function diff($date1, $date2) {
    $diff = abs(strtotime($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));
    return $days;
}
?>

SamGoody · Answer 10 · 2011-07-21T11:09:24.760

PHP has methods for working with Unix timestamps.

As has been noted by others, by working with seconds since the Unix date, it is easy to calculate times.

PHP's strtotime() converts a date to a timestamp:

$diff = round((strtotime($list['start']) - strtotime($list['finish'])) / 86400);

If you wish to calculate till the current time, time() provides the timestamp of "now":

$diff = round((time() - strtotime($list['date'])) / 86400);

86400 is the number of seconds in a day.
If you wish to convert to years use 31557000, which is almost exactly 365.24219 * 86400.

An added advantage here is that strtotime can take the input date in almost any human readable format, so it is very easy to work with within the code.

What can use for DateTime::diff() for PHP 5.2?

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