PHP date - Unix timestamp

Question

Could somebody please help me to resolve this PHP timestamp riddle

I'm trying to find number of weekdays based on two given dates.

I'm getting a string date eg starting date: 01/10/2019 (1st of October 2019) and explode it to a array => $s

Array $s set like this:

Array ( [0] => 01 [1] => 10 [2] => 2019 )

then I made a Unix timestamp from it :

date('U', mktime(0, 0, 0, $s[1], $s[0], $s[2]));

I get 1569884400 as the result of above date conversion.

then I use a loop to run this and so I can find number of week days based on the 2 given dates..

this is a result of that loop and as you can see for some reason 27th October repeating.

Just to clarify the top of this PHP page i have timezone set

date_default_timezone_set('Europe/London');

this is the code behind the below result

total $total_days based on last date (in this case 31st October 2019 and ie: 31 days)

October 2019 should have 23 working days, Since I get 2x 27th Octobers I will get only 22 weekdays.

What am I doing wrong here?

 $wk_days = array(1, 2, 3, 4, 5);
  $days = 0;
  for ($i = 0; $i < $total_days; $i++) {
    $tmp = $first + ($i * 86400);
    echo  $i." x86400 add to ". $first . " ==> " . $tmp. " and the new date is "  .date('Y-m-d',$tmp) ."<br/>";

    $chk = date('w', $tmp);

    if (in_array($chk, $wk_days)) {
      $days++;
    }
  }

0 x86400 add to 1569884400 ==> 1569884400 and the new date is 2019-10-01

1 x86400 add to 1569884400 ==> 1569970800 and the new date is 2019-10-02

2 x86400 add to 1569884400 ==> 1570057200 and the new date is 2019-10-03

3 x86400 add to 1569884400 ==> 1570143600 and the new date is 2019-10-04

4 x86400 add to 1569884400 ==> 1570230000 and the new date is 2019-10-05

...

23 x86400 add to 1569884400 ==> 1571871600 and the new date is 2019-10-24

24 x86400 add to 1569884400 ==> 1571958000 and the new date is 2019-10-25

25 x86400 add to 1569884400 ==> 1572044400 and the new date is 2019-10-26

26 x86400 add to 1569884400 ==> 1572130800 and the new date is 2019-10-27 //***

27 x86400 add to 1569884400 ==> 1572217200 and the new date is 2019-10-27 //***

28 x86400 add to 1569884400 ==> 1572303600 and the new date is 2019-10-28

29 x86400 add to 1569884400 ==> 1572390000 and the new date is 2019-10-29

30 x86400 add to 1569884400 ==> 1572476400 and the new date is 2019-10-30

1569884400 is September 30th, not October 1st. Could that be the problem? (Note that I haven't really gone through the rest of the logic in detail.) I think first and foremost you should double-check your conversion at the beginning. — Andrew
– Andrew, Commented Jan 15, 2020 at 13:59
You compute the date at midnight. You shoud compute it at 12 pm in order to avoid having a problem when the time changes from summer to winter on the last Sunday of October. So use date('U', mktime(12, 0, 0, $s[1], $s[0], $s[2])); — Pierre François
– Pierre François, Commented Jan 15, 2020 at 14:00

Dharman · Accepted Answer · 2020-01-15 15:17:36Z

2

You don't need anything that complicated. You can have a simple while loop and count the weekdays. It might not be the most efficient way to do it, but it should work.

function getNoOfWeekdays(string $startdate, string $enddate, string $format = 'd/m/Y'): int
{
    $start = date_create_from_format($format, $startdate);
    $end = date_create_from_format($format, $enddate);

    $count = 0;
    while ($start <= $end) {
        if ($start->format('N') < 6) $count++;
        $start->modify('+1 days');
    }

    return $count;
}

answered Jan 15, 2020 at 15:17

Dharman♦

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1 Comment

jspit Over a year ago

Instead of a simple while loop, this can also be done with DatePeriod and a DateInterval "+1 weekday". The start date must be modified for this.

Matías Rodríguez · Accepted Answer · 2020-01-15 14:07:26Z

1

Maybe you could try with this code:

  $initial = "2019-10-01";
  $last = "2019-10-31";
  $wk_days = array(1, 2, 3, 4, 5);
  $days = 0;
  $date = $initial;
  do{

    $chk = date("w",strtotime($date));
    if(in_array($chk,$wk_days))
      $days++;
    $date = date("Y-m-d",strtotime("$date +1 day"));

  }while($date!=date("Y-m-d",strtotime("$last +1 day")));
    echo $days;
?>

answered Jan 15, 2020 at 14:07

Matías Rodríguez

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jspit · Accepted Answer · 2020-01-15 14:40:26Z

1

I created a PHP extension for DateTime API called dt. You can find it here. Using it is very easy:

$start = '2019-10-01';
$end = '2019-11-01'; 

$start = dt::create($start);

//exclude end with true as 3.parameter
$countWeekdaysOct2019 = $start->countDaysTo('1,2,3,4,5',$end, true); //23

$countSundaysOct2019 = $start->countDaysTo('0',$end, true); //4

$countSaturdaysOct2019 = $start->countDaysTo('6',$end, true); //4

answered Jan 15, 2020 at 14:40

jspit

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Comments

Pierre François · Accepted Answer · 2020-01-16 09:36:53Z

1

Try this (see my comment above) in order to avoid to problem of changing from summer to winter time:

$first = date('U', mktime(12 /* 12 pm!! */, 0, 0, $s[1], $s[0], $s[2]));
//
// some code here that the author didn't show and we hardly can extrapolate
// ...
//
$wk_days = array(1, 2, 3, 4, 5);
  $days = 0;
  for ($i = 0; $i < $total_days; $i++) {
    $tmp = $first + ($i * 86400);
    echo  $i." x86400 add to ". $first . " ==> " . $tmp. " and the new date is "  .date('Y-m-d',$tmp) ."<br/>";

    $chk = date('w', $tmp);

    if (in_array($chk, $wk_days)) {
      $days++;
    }
  }

edited Jan 16, 2020 at 9:36

answered Jan 15, 2020 at 14:06

Pierre François

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4 Comments

sam6640621 Over a year ago

Does this mean we have to manually handle summer time although we introduce the time zone?

Pierre François Over a year ago

No, no, what I mean is that if you compute the date from the timestamp at the border of your day (at midnight), and the border moves twice in a year, with a day of 25 hours in autumn and a day of 23 hours in the spring, you will create unwanted side effects. That is the reason why I suggest you to compute the timestamp of the day at 12 pm, instead of midnight.

Pierre François Over a year ago

After a while, I realize that what may work in the time zone of London may not work at the antipods. It should me more prudent to perform date arithmetics in a way that doesn't interact with timezones and daylight saving times.

sam6640621 Over a year ago

I updated the code as you suggest and it work well. Thanks for the prompt help.

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