Get week number (in the year) from a date PHP

Question

I want to take a date and work out its week number.

So far, I have the following. It is returning 24 when it should be 42.

Answer 1

The rule is that the first week of a year is the week that contains the first Thursday of the year.

I personally use Zend_Date for this kind of calculation and to get the week for today is this simple. They have a lot of other useful functions if you work with dates.

$now = Zend_Date::now();
$week = $now->get(Zend_Date::WEEK);
// 10

Answer 2

To get the week number for a date in North America I do like this:

function week_number($n)
{
    $w = date('w', $n);
    return 1 + date('z', $n + (6 - $w) * 24 * 3600) / 7;
}

$n = strtotime('2022-12-27');
printf("%s: %d\n", date('D Y-m-d', $n), week_number($n));

and get:

Tue 2022-12-27: 53

Answer 3

I have tried to solve this question for years now, I thought I found a shorter solution but had to come back again to the long story. This function gives back the right ISO week notation:

/**
 * calcweek("2018-12-31") => 1901
 * This function calculates the production weeknumber according to the start on 
 * monday and with at least 4 days in the new year. Given that the $date has
 * the following format Y-m-d then the outcome is and integer.
 *
 * @author M.S.B. Bachus
 *
 * @param date-notation PHP "Y-m-d" showing the data as yyyy-mm-dd
 * @return integer
 **/
function calcweek($date) {
  // 1. Convert input to $year, $month, $day
  $dateset      = strtotime($date);
  $year         = date("Y", $dateset);
  $month        = date("m", $dateset);
  $day          = date("d", $dateset);

  $referenceday = getdate(mktime(0,0,0, $month, $day, $year));
  $jan1day      = getdate(mktime(0,0,0,1,1,$referenceday[year]));

  // 2. check if $year is a  leapyear
  if ( ($year%4==0 && $year%100!=0) || $year%400==0) {
    $leapyear = true;
  } else {
    $leapyear = false;
  }

  // 3. check if $year-1 is a  leapyear
  if ( (($year-1)%4==0 && ($year-1)%100!=0) || ($year-1)%400==0 ) {
    $leapyearprev = true;
  } else {
    $leapyearprev = false;
  }

  // 4. find the dayofyearnumber for y m d
  $mnth = array(0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334);
  $dayofyearnumber = $day + $mnth[$month-1];
  if ( $leapyear && $month > 2 ) { $dayofyearnumber++; }

  // 5. find the jan1weekday for y (monday=1, sunday=7)
  $yy = ($year-1)%100;
  $c  = ($year-1) - $yy;
  $g  = $yy + intval($yy/4);
  $jan1weekday = 1+((((intval($c/100)%4)*5)+$g)%7);

  // 6. find the weekday for y m d
  $h = $dayofyearnumber + ($jan1weekday-1);
  $weekday = 1+(($h-1)%7);

  // 7. find if y m d falls in yearnumber y-1, weeknumber 52 or 53
  $foundweeknum = false;
  if ( $dayofyearnumber <= (8-$jan1weekday) && $jan1weekday > 4 ) {
    $yearnumber = $year - 1;
    if ( $jan1weekday = 5 || ( $jan1weekday = 6 && $leapyearprev )) {
      $weeknumber = 53;
    } else {
      $weeknumber = 52;
    }
    $foundweeknum = true;
  } else {
    $yearnumber = $year;
  }

  // 8. find if y m d falls in yearnumber y+1, weeknumber 1
  if ( $yearnumber == $year && !$foundweeknum) {
    if ( $leapyear ) {
      $i = 366;
    } else {
      $i = 365;
    }
    if ( ($i - $dayofyearnumber) < (4 - $weekday) ) {
      $yearnumber = $year + 1;
      $weeknumber = 1;
      $foundweeknum = true;
    }
  }

  // 9. find if y m d falls in yearnumber y, weeknumber 1 through 53
  if ( $yearnumber == $year && !$foundweeknum ) {
    $j = $dayofyearnumber + (7 - $weekday) + ($jan1weekday - 1);
    $weeknumber = intval( $j/7 );
    if ( $jan1weekday > 4 ) { $weeknumber--; }
  }

  // 10. output iso week number (YYWW)
  return ($yearnumber-2000)*100+$weeknumber;
}

I found out that my short solution missed the 2018-12-31 as it gave back 1801 instead of 1901. So I had to put in this long version which is correct.

Answer 4

Becomes more difficult when you need year and week.
Try to find out which week is 01.01.2017.
(It is the 52nd week of 2016, which is from Mon 26.12.2016 - Sun 01.01.2017).

After a longer search I found

strftime('%G-%V',strtotime("2017-01-01"))

Result: 2016-52

---

https://www.php.net/manual/de/function.strftime.php
ISO-8601:1988 week number of the given year, starting with the first week of the year with at least 4 weekdays, with Monday being the start of the week. (01 through 53)

---

The equivalent in mysql is DATE_FORMAT(date, '%x-%v') https://www.w3schools.com/sql/func_mysql_date_format.asp
Week where Monday is the first day of the week (01 to 53).

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Could not find a corresponding solution with date() or DateTime.
At least not without solutions like "+1day, last monday".

Answer 5

Use PHP's date function
http://php.net/manual/en/function.date.php

date("W", $yourdate)

Answer 6

function last_monday($date) 
{
    if (!is_numeric($date))
        $date = strtotime($date);
    if (date('w', $date) == 1)
        return $date;
    else
        return date('Y-m-d',strtotime('last monday',$date));
}
$date = '2021-01-04';  //Enter custom date
$year = date('Y',strtotime($date));
$date1 = new DateTime($date);
$ldate = last_monday($year."-01-01");
$date2 = new DateTime($ldate);
$diff = $date2->diff($date1)->format("%a");
$diff = $diff/7;
$week = intval($diff) + 1;
echo $week;
//Returns 2.