How to know week number from date in php?

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.

Today, I would like to show you how to retrieve the calender week (week number) to an arbitrary given date with PHP.

This is much easier than you might suspect: we only need the function date() and the formatting character "W" for it. When passing this character as a first parameter and a date as a second parameter to the date() function, we get the calendar week respectively the week number of the year for our date as a result.

$dat = time();         // current date

echo date("W", $dat);  // output, for example: 03 or 10

According to ISO-8601, the week begins at Monday for PHP and the output is given with two digits as string. If we want to have a single digit number for the week numbers 1 to 9, we can just write (int)date("W", $dat) to get it. With this, the result of the function is converted from a string to an integer number whose output is naturally single-digit without any leading zeros.

Year of the Calendar Week

There is one problem at each turn of the year. The problem can occur whenever the first calendar week of the new year is already starting in December or if the last calendar week of a year is going over into January. In both of this cases, the usage of date("W Y", $date) leads to wrong results:

$dat = strtotime('2016-01-01');

echo date("W Y", $dat);  // output: 53 2016 

This example shows, what is going wrong: for the date 2016-01-01, we get the 53th week number of 2016. The reason is that the "Y" is referring to the year number of the passed date - this is 2016. However, we need 2015 as reside, because January, 1 is belonging to the old year week in this case from the point of view of the calendar week.

Fortunately, the developers of PHP have thought of this special case and bestowed the parameter "o" (lowercase O) upon us. If we use this parameter instead of the usual parameter Y for the year with the date() function, we will receive the year suitable for the passed date's week number. Here is an example:

$dat = strtotime('2016-01-01');

echo date("W o", $dat);  // output: 53 2015

Using this parameter, the result is correct: we get the 53th calendar week of 2015.

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