How to count days between two dates in PHP?

Question

If I have a couple of strings $startDate and $endDate which are set to (for instance) "2011/07/01" and "2011/07/17" (meaning 1 July 2011 and 17 July 2011). How would I count the days from start date to end date? In the example given, it would be 17 days.

see http://stackoverflow.com/questions/2040560/finding-the-number-of-days-between-two-dates — Eric Lavoie, Jan 04 '16 at 20:16

score 54 · Accepted Answer · answered Sep 06 '10 at 19:46

54

Here is the raw way to do it

$startTimeStamp = strtotime("2011/07/01");
$endTimeStamp = strtotime("2011/07/17");

$timeDiff = abs($endTimeStamp - $startTimeStamp);

$numberDays = $timeDiff/86400;  // 86400 seconds in one day

// and you might want to convert to integer
$numberDays = intval($numberDays);

answered Sep 06 '10 at 19:46

axsuul

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Will not work correctly when you have local dates that include a dayling savings time switchover, because on those dates you have days with 23 or 25 hours. – Michael Borgwardt Sep 06 '10 at 19:53
Actually, it says 16 days for those dates.. Do I need to make dates ignore daylight saving time..? – cannyboy Sep 06 '10 at 22:43
It should be 16 days because July 1 and July 17 are 16 days apart whereas July 1 and July 18 are 17 days apart – axsuul Sep 06 '10 at 23:09
@MichaelBorgwardt you have two solid dates, I don't see how daylight saving can affect this. – pronebird Mar 19 '14 at 22:30
2

@Andy: if by "solid" you mean "literal", sure, but the question says "for instance", not "I need a solution for these specific dates in 2011 and don't care for any problems that might arise with other dates". – Michael Borgwardt Mar 19 '14 at 22:42
@MichaelBorgwardt approved. – pronebird Mar 20 '14 at 11:53
They say it's better to use the `Datetime` class, since `strtotime` can't compare dates further than 2038 year ... – Pathros Jun 02 '17 at 18:36

wuputah · Answer 2 · 2010-09-06T19:41:37.840

35

Use DateTime::diff (aka date_diff):

$datetime1 = new DateTime('2009-10-11');
$datetime2 = new DateTime('2009-10-13');
$interval = $datetime1->diff($datetime2);

Or:

$datetime1 = date_create('2009-10-11');
$datetime2 = date_create('2009-10-13');
$interval = date_diff($datetime1, $datetime2);

You can then get the interval as a integer by calling $interval->days.

edited Sep 06 '10 at 19:41

answered Sep 06 '10 at 19:31

wuputah

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That doesn't seem to work with my initial strings of "2011/07/01" and "2011/07/17" ... "Call to undefined function date_diff().." . I'm using PHP 5.2.10 – cannyboy Sep 06 '10 at 21:40
4

Per the documentation (linked at the top), this is PHP >= 5.3.0. For next time, please include the PHP version in your question. – wuputah Sep 06 '10 at 23:13
does only work if you don't have a time in your Date Times otherwise it may return 1-2 days to few – Dukeatcoding Jan 22 '14 at 14:09
2

bug : $interval->days return 6015 days ! And if i use the $interval->format("%d"), it only return the differences between the days, without taking into account the months and years. – Alex Sep 02 '14 at 14:54
1

Note that you need to add 1 to the result if you want to include both the start and end day in the count. – Andrew Downes Nov 18 '17 at 08:53

score 5 · Answer 3 · answered Sep 06 '10 at 19:30

5

PHP has a date_diff() function to do this.

answered Sep 06 '10 at 19:30

Paul Schreiber

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score 3 · Answer 4 · answered May 31 '13 at 16:41

3

<?php
$datetime1 = new DateTime('2009-10-11');
$datetime2 = new DateTime('2009-10-13');
$interval = $datetime1->diff($datetime2);
echo $interval->format('%R%a days');
?>

Source: http://www.php.net/manual/en/datetime.diff.php

answered May 31 '13 at 16:41

Layke

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Dukeatcoding · Answer 5 · 2014-01-22T14:31:24.203

In case your DateTime has also hour:minutes:seconds and you still want to have the number of days..

   /**
     * Returns the total number of days between to DateTimes, 
     * if it is within the same year
     * @param $start
     * @param $end
     */
    public function dateTimesToDays($start,$end){
       return intval($end->format('z')) - intval($start->format('z')) + 1;
    }

https://github.com/dukeatcoding/timespan-converter

score 3 · Answer 6 · edited Nov 07 '17 at 06:32

3

$date1 = date_create("2017-04-15");
$date2 = date_create("2017-05-18");

//difference between two dates
$diff = date_diff($date1,$date2);

//count days
echo 'Days Count - '.$diff->format("%a");

edited Nov 07 '17 at 06:32

Neverever

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answered Nov 07 '17 at 04:58

khoir

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score 1 · Answer 7 · edited Oct 09 '12 at 17:04

1

None of the solutions worked for me. For those who are still on PHP 5.2 (DateTime::diff was introduced in 5.3), this solution works:

function howDays($from, $to) {
    $first_date = strtotime($from);
    $second_date = strtotime($to);
    $offset = $second_date-$first_date; 
    return floor($offset/60/60/24);
}

edited Oct 09 '12 at 17:04

sobering

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answered Nov 16 '11 at 15:00

Tomás Crespo García

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Jose María Simón · Answer 8 · 2014-11-26T20:28:50.880

If you want to know the number of days (if any), the number of hours (if any), minutues (if any) and seconds, you can do the following:

$previousTimeStamp = strtotime("2011/07/01 21:12:34");
$lastTimeStamp = strtotime("2013/09/17 12:34:11");

$menos=$lastTimeStamp-$previousTimeStamp;

$mins=$menos/60;
if($mins<1){
$showing= $menos . " seconds ago";
}
else{
$minsfinal=floor($mins);
$secondsfinal=$menos-($minsfinal*60);
$hours=$minsfinal/60;
if($hours<1){
$showing= $minsfinal . " minutes and " . $secondsfinal. " seconds ago";

}
else{
$hoursfinal=floor($hours);
$minssuperfinal=$minsfinal-($hoursfinal*60);
$days=$hoursfinal/24;
if($days<1){
$showing= $hoursfinal . "hours, " . $minssuperfinal . " minutes and " . $secondsfinal. " seconds ago";

}
else{
$daysfinal=floor($days);
$hourssuperfinal=$hoursfinal-($daysfinal*24);
$showing= $daysfinal. "days, " .$hourssuperfinal . " hours, " . $minssuperfinal . " minutes and " . $secondsfinal. " seconds ago";
}}}

echo $showing;

You could use the same logic if you want to add months and years.

score 0 · Answer 9 · edited Feb 21 '15 at 18:10

0

Simple way to count is,

$currentdate = date('Y-m-d H:i:s');
$after1yrdate =  date("Y-m-d H:i:s", strtotime("+1 year", strtotime($data)));

$diff = (strtotime($after1yrdate) - strtotime($currentdate)) / (60 * 60 * 24);

echo "<p style='color:red'>The difference is ".round($diff)." Days</p>";

edited Feb 21 '15 at 18:10

AbcAeffchen

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answered Feb 21 '15 at 17:51

Archa

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score 0 · Answer 10 · answered Dec 11 '18 at 09:57

i created a function in which if you pass two dates than it will return day wise value. For better understanding please see the output of start date : 2018-11-12 11:41:19 and End Date 2018-11-16 12:07:26

private function getTimeData($str1,$str2){
$datetime1 = strtotime($str1);
$datetime2 = strtotime($str2);
$myArray = array();
if(date('d', $datetime2) != date('d', $datetime1) ||  date('m', $datetime2) != date('m', $datetime1) || date('y', $datetime2) != date('y', $datetime1)){
    $exStr1 = explode(' ',$str1);
    $exStr2 = explode(' ',$str2);
    $datediff = strtotime($exStr2[0]) - strtotime($exStr1[0]);
    $totalDays = round($datediff / (60 * 60 * 24));
    $actualDate1 = $datetime1;
    $actualDate2 = date('Y-m-d', $datetime1)." 23:59:59";
    $interval  = abs(strtotime($actualDate2)-$actualDate1);
    $minutes   = round($interval / 60);
    $myArray[0]['startDate'] = date('Y-m-d H:i:s', $actualDate1); 
    $myArray[0]['endDate'] = $actualDate2;
    $myArray[0]['minutes'] = $minutes;
    $i = 1;
    if($totalDays > 1){    
        for($i=1; $i<$totalDays; $i++){
            $dayString = "+".$i." day";
            $edate = strtotime($dayString, $actualDate1);
            $myArray[$i]['startDate'] = date('Y-m-d', $edate)." 00:00:00"; 
            $myArray[$i]['endDate'] = date('Y-m-d', $edate)." 23:59:59"; 
            $myArray[$i]['minutes'] = 1440;
        }
    }
    $actualSecDate1 = date('Y-m-d', $datetime2)." 00:00:00";
    $actualSecDate2 = $datetime2;
    $interval  = abs(strtotime($actualSecDate1)-$actualSecDate2);
    $minutes   = round($interval / 60);
    $myArray[$i]['startDate'] = $actualSecDate1; 
    $myArray[$i]['endDate'] = date('Y-m-d H:i:s', $actualSecDate2); 
    $myArray[$i]['minutes'] = $minutes;
}
else{
    $interval  = abs($datetime2-$datetime1);
    $minutes   = round($interval / 60);
    $myArray[0]['startDate'] = date('Y-m-d H:i:s', $datetime1); 
    $myArray[0]['endDate'] = date('Y-m-d H:i:s', $datetime2);
    $myArray[0]['minutes'] = $minutes;
}

return $myArray;
}

Output

Array
(
[0] => Array
    (
        [startDate] => 2018-11-12 11:41:19
        [endDate] => 2018-11-12 23:59:59
        [minutes] => 739
    )

[1] => Array
    (
        [startDate] => 2018-11-13 00:00:00
        [endDate] => 2018-11-13 23:59:59
        [minutes] => 1440
    )

[2] => Array
    (
        [startDate] => 2018-11-14 00:00:00
        [endDate] => 2018-11-14 23:59:59
        [minutes] => 1440
    )

[3] => Array
    (
        [startDate] => 2018-11-15 00:00:00
        [endDate] => 2018-11-15 23:59:59
        [minutes] => 1440
    )

[4] => Array
    (
        [startDate] => 2018-11-16 00:00:00
        [endDate] => 2018-11-16 12:07:26
        [minutes] => 727
    )
)

score -1 · Answer 11 · edited Jul 14 '17 at 09:17

-1

You can use date_diff to calculate the difference between two dates:

$date1 = date_create("2013-03-15");
$date2 = date_create("2013-12-12");
$diff =  date_diff($date1 , $date2);
echo  $diff->format("%R%a days");

edited Jul 14 '17 at 09:17

Peter

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answered Jul 14 '17 at 08:32

Anuj Chouadhary

1

How to count days between two dates in PHP?

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