Get week of month php

I have an array of random dates (not coming from MySQL). I need to group them by the week as Week1, Week2, and so on upto Week5.

What I have is this:

$dates = array('2015-09-01','2015-09-05','2015-09-06','2015-09-15','2015-09-17');

What I need is a function to get the week number of the month by providing the date.

I know that I can get the weeknumber by doing date('W',strtotime('2015-09-01')); but this week number is the number between year (1-52) but I need the week number of the month only, e.g. in Sep 2015 there are 5 weeks:

• Week1 = 1st to 5th
• Week2 = 6th to 12th
• Week3 = 13th to 19th
• Week4 = 20th to 26th
• Week5 = 27th to 30th

I should be able to get the week Week1 by just providing the date e.g.

$weekNumber = getWeekNumber('2015-09-01') //output 1;
$weekNumber = getWeekNumber('2015-09-17') //output 3;

Anders

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asked Sep 16, 2015 at 18:18

1

I think this relationship should be true and come in handy:

Week of the month = Week of the year - Week of the year of first day of month + 1

We also need to make sure that "overlapping" weeks from the previous year are handeled correctly - if January 1st is in week 52 or 53, it should be counted as week 0. In a similar fashion, if a day in December is in the first week of the next year, it should be counted as 53. (Previous versions of this answer failed to do this properly.)

 51) {
        // It's the last week of the previos year.
        return 0;
    }
    else if (date('n', $date) == "12" && $weekOfYear == 1) {
        // It's the first week of the next year.
        return 53;
    }
    else {
        // It's a "normal" week.
        return $weekOfYear;
    }
}

// A few test cases.
echo weekOfMonth(strtotime("2020-04-12")) . " "; // 2
echo weekOfMonth(strtotime("2020-12-31")) . " "; // 5
echo weekOfMonth(strtotime("2020-01-02")) . " "; // 1
echo weekOfMonth(strtotime("2021-01-28")) . " "; // 5
echo weekOfMonth(strtotime("2018-12-31")) . " "; // 6

To get weeks that starts with sunday, simply replace date("W", ...) with strftime("%U", ...).

answered Sep 17, 2015 at 7:29

AndersAnders

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8

You can use the function below, fully commented:

/**
 * Returns the number of week in a month for the specified date.
 *
 * @param string $date
 * @return int
 */
function weekOfMonth($date) {
    // estract date parts
    list($y, $m, $d) = explode('-', date('Y-m-d', strtotime($date)));
    
    // current week, min 1
    $w = 1;
    
    // for each day since the start of the month
    for ($i = 1; $i < $d; ++$i) {
        // if that day was a sunday and is not the first day of month
        if ($i > 1 && date('w', strtotime("$y-$m-$i")) == 0) {
            // increment current week
            ++$w;
        }
    }
    
    // now return
    return $w;
}

answered Sep 17, 2015 at 6:52

Matteo TassinariMatteo Tassinari

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4

The corect way is

function weekOfMonth($date) {
    $firstOfMonth = date("Y-m-01", strtotime($date));
    return intval(date("W", strtotime($date))) - intval(date("W", strtotime($firstOfMonth)));
}

answered Nov 18, 2015 at 6:38

GoendgGoendg

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1

I have created this function on my own, which seems to work correctly. In case somebody else have a better way of doing this, please share.. Here is what I have done.

function weekOfMonth($qDate) {
    $dt = strtotime($qDate);
    $day  = date('j',$dt);
    $month = date('m',$dt);
    $year = date('Y',$dt);
    $totalDays = date('t',$dt);
    $weekCnt = 1;
    $retWeek = 0;
    for($i=1;$i<=$totalDays;$i++) {
        $curDay = date("N", mktime(0,0,0,$month,$i,$year));
        if($curDay==7) {
            if($i==$day) {
                $retWeek = $weekCnt+1;
            }
            $weekCnt++;
        } else {
            if($i==$day) {
                $retWeek = $weekCnt;
            }
        }
    }
    return $retWeek;
}

echo weekOfMonth('2015-09-08') // gives me 2;

answered Sep 17, 2015 at 7:22

Asif HussainAsif Hussain

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2

function getWeekOfMonth(DateTime $date) {
    $firstDayOfMonth = new DateTime($date->format('Y-m-1'));

    return ceil(($firstDayOfMonth->format('N') + $date->format('j') - 1) / 7);
}

Goendg solution does not work for 2016-10-31.

answered Jun 10, 2016 at 16:48

j4r3kj4r3k

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function weekOfMonth($strDate) {
  $dateArray = explode("-", $strDate);
  $date = new DateTime();
  $date->setDate($dateArray[0], $dateArray[1], $dateArray[2]);
  return floor((date_format($date, 'j') - 1) / 7) + 1;  
}

weekOfMonth ('2015-09-17') // returns 3

answered Jun 7, 2018 at 7:10

1

Given the time_t wday (0=Sunday through 6=Saturday) of the first of the month in firstWday, this returns the (Sunday-based) week number within the month:

weekOfMonth = floor((dayOfMonth + firstWday - 1)/7) + 1 

Translated into PHP:

function weekOfMonth($dateString) {
  list($year, $month, $mday) = explode("-", $dateString);
  $firstWday = date("w",strtotime("$year-$month-1"));
  return floor(($mday + $firstWday - 1)/7) + 1;
}

answered Apr 26, 2016 at 2:58

Mark ReedMark Reed

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You can also use this simple formula for finding week of the month

$currentWeek = ceil((date("d",strtotime($today_date)) - date("w",strtotime($today_date)) - 1) / 7) + 1;

ALGORITHM :

Date = '2018-08-08' => Y-m-d

1. Find out day of the month eg. 08
2. Find out Numeric representation of the day of the week minus 1 (number of days in week) eg. (3-1)
3. Take difference and store in result
4. Subtract 1 from result
5. Divide it by 7 to result and ceil the value of result
6. Add 1 to result eg. ceil(( 08 - 3 ) - 1 ) / 7) + 1 = 2

answered Aug 8, 2018 at 10:38

My function. The main idea: we would count amount of weeks passed from the month's first date to current. And the current week number would be the next one. Works on rule: "Week starts from monday" (for sunday-based type we need to transform the increasing algorithm)

function GetWeekNumberOfMonth ($date){
    echo $date -> format('d.m.Y');
    //define current year, month and day in numeric
    $_year = $date -> format('Y');
    $_month = $date -> format('n');
    $_day = $date -> format('j');
    $_week = 0; //count of weeks passed
    for ($i = 1; $i < $_day; $i++){
        echo "\n\n-->";
        $_newDate = mktime(0,0,1, $_month, $i, $_year);
        echo "\n";
        echo date("d.m.Y", $_newDate);
        echo "-->";
        echo date("N", $_newDate);
        //on sunday increasing weeks passed count
        if (date("N", $_newDate) == 7){
            echo "New week";
            $_week += 1;
        }

    }
    return $_week + 1; // as we are counting only passed weeks the current one would be on one higher
}

$date = new DateTime("2019-04-08");
echo "\n\nResult: ". GetWeekNumberOfMonth($date);

answered May 6, 2019 at 0:04

$month = 6;
$year = 2021;           
$week = date("W", strtotime($year . "-" . $month ."-01"));

$str='';
$str .= date("d-m-Y", strtotime($year . "-" . $month ."-01")) ."to";
$unix = strtotime($year."W".$week ."+1 week");
while(date("m", $unix) == $month){
 $str .= date("d-m-Y", $unix-86400) . "|";
 $str .= date("d-m-Y", $unix) ."to"; 
 $unix = $unix + (86400*7);
}
$str .= date("d-m-Y", strtotime("last day of ".$year . "-" . $month));

$weeks_ar = explode('|',$str);
echo '
'; print_r($weeks_ar);
working fine.
 answered Jun 25, 2021 at 8:29
 
    // Current week of the month starts with Sunday
 
    $first_day_of_the_week = 'Sunday';
    $start_of_the_week1    = strtotime("Last $first_day_of_the_week");     
    
    if (strtolower(date('l')) === strtolower($first_day_of_the_week)) {
         $start_of_the_week1 = strtotime('today');
     }
     $end_of_the_week1   = $start_of_the_week1 + (60 * 60 * 24 * 7) - 1;

// Get the date format
print date('Y-m-d', $start_of_the_week1) . ' 00:00:00';
print date('Y-m-d', $end_of_the_week1) . ' 23:59:59';
 answered Jul 16, 2021 at 16:21
 
BalajiBalaji
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// self::DAYS_IN_WEEK = 7;
function getWeeksNumberOfMonth(): int
{
    $currentDate            = new \DateTime();
    $dayNumberInMonth       = (int) $currentDate->format('j');
    $dayNumberInWeek        = (int) $currentDate->format('N');
    $dayNumberToLastSunday  = $dayNumberInMonth - $dayNumberInWeek;
    $daysCountInFirstWeek   = $dayNumberToLastSunday % self::DAYS_IN_WEEK;
    $weeksCountToLastSunday = ($dayNumberToLastSunday - $daysCountInFirstWeek) / self::DAYS_IN_WEEK;

    $weeks = [];
    array_push($weeks, $daysCountInFirstWeek);
    for ($i = 0; $i < $weeksCountToLastSunday; $i++) {
        array_push($weeks, self::DAYS_IN_WEEK);
    }
    array_push($weeks, $dayNumberInWeek);

    if (array_sum($weeks) !== $dayNumberInMonth) {
        throw new Exception('Logic is not valid');
    }

    return count($weeks);
}
Short variant:
(int) (new \DateTime())->format('W') - (int) (new \DateTime('first day of this month'))->format('W') + 1;
 answered Sep 26, 2018 at 14:33
 
3
There is a many solutions but here is one my
solution that working well in the most cases.
function current_week ($date = NULL) {
    if($date) {
        if(is_numeric($date) && ctype_digit($date) && strtotime(date('Y-m-d H:i:s',$date)) === (int)$date)
            $unix_timestamp = $date;
        else
            $unix_timestamp = strtotime($date);
    } else $unix_timestamp = time();

    return (ceil((date('d', $unix_timestamp) - date('w', $unix_timestamp) - 1) / 7) + 1);
}
It accept unix timestamp, normal date or return current week from the time() if you not pass any value.
Enjoy!
 answered Sep 3, 2019 at 13:45
 
I know this an old post but i have an idea!
$datetime0 = date_create("1970-01-01");
$datetime1 = date_create(date("Y-m-d",mktime(0,0,0,$m,"01",$Y)));
$datetime2 = date_create(date("Y-m-d",mktime(0,0,0,$m,$d,$Y)));

$interval1 = date_diff($datetime0, $datetime1);
$daysdiff1= $interval1->format('%a');

$interval2 = date_diff($datetime0, $datetime2);
$daysdiff2= $interval2->format('%a');

$week1=round($daysdiff1/7);
$week2=round($daysdiff2/7);

$WeekOfMonth=$week2-$week1+1;
 answered Dec 2, 2019 at 15:25

1
$date = new DateTime('first Monday of this month');
$thisMonth = $date->format('m');
$mondays_arr = [];

// Get all the Mondays in the current month and store in array
while ($date->format('m') === $thisMonth) {
    //echo $date->format('Y-m-d'), "\n";
    $mondays_arr[] = $date->format('d');
    $date->modify('next Monday');
}

// Get the day of the week (1-7 from monday to sunday)
$day_of_week = date('N') - 1;

// Get the day of month (1 to 31) 
$current_week_monday_date = date('j') - $day_of_week;

/*$day_of_week = date('N',mktime(0, 0, 0, 2, 11, 2020)) - 1;
$current_week_monday_date = date('j',mktime(0, 0, 0, 2, 11, 2020)) - $day_of_week;*/

$week_no = array_search($current_week_monday_date,$mondays_arr) + 1;
echo "Week No: ". $week_no;
 answered Feb 11, 2020 at 8:26
 
How about this function making use of PHP's relative dates? This function assumes the week ends on Saturday. But this can be changed easily.
function get_weekNumMonth($date) {
    $CI = &get_instance();
    $strtotimedate = strtotime($date);
    $firstweekEnd = date('j', strtotime("FIRST SATURDAY OF " . date("F", $strtotimedate) . " " . date("Y", $strtotimedate)));
    $cutoff = date('j', strtotime($date));
    $weekcount = 1;
    while ($cutoff > $firstweekEnd) {
        $weekcount++;
        $firstweekEnd += 7; // move to next week
    }
    return $weekcount;
}
 answered Feb 17, 2021 at 16:27

Chin. UdaraChin. Udara
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This function returns the integer week number of the current month. Weeks always start on
Monday and counting always starts with 1.
function weekOfmonth(DateTime $date)
{
  $dayFirstMonday = date_create('first monday of '.$date->format('F Y'))->format('j');
  return (int)(($date->format('j') - $dayFirstMonday +7)/7) + ($dayFirstMonday == 1 ? 0 : 1);
}
Example of use
echo weekOfmonth(new DateTime("2020-04-12"));  //2
A test for all days from 1900-2038 with the accepted solution from @Anders as a reference:
//reference functions
//integer $date (Timestamp) 
function weekOfMonthAnders($date) {
    //Get the first day of the month.
    $firstOfMonth = strtotime(date("Y-m-01", $date));
    //Apply above formula.
    return weekOfYear($date) - weekOfYear($firstOfMonth) + 1;
}

function weekOfYear($date) {
    $weekOfYear = intval(date("W", $date));
    if (date('n', $date) == "1" && $weekOfYear > 51) {
        // It's the last week of the previos year.
        return 0;
    }
    else if (date('n', $date) == "12" && $weekOfYear == 1) {
        // It's the first week of the next year.
        return 53;
    }
    else {
        // It's a "normal" week.
        return $weekOfYear;
    }
}

//this function
function weekOfmonth(DateTime $date)
{
  $dayFirstMonday = date_create('first monday of '.$date->format('F Y'))->format('j');
  return (int)(($date->format('j') - $dayFirstMonday +7)/7) + ($dayFirstMonday == 1 ? 0 : 1);
}

$dt = date_create('1900-01-01');
$end = date_create('2038-01-02');
$countOk = 0;
$countError = 0;
for(;$dt < $end; $dt->modify('+1 Day')){
    $ts = $dt->getTimestamp();
    if(weekOfmonth($dt) === weekOfMonthAnders($ts)){
       ++$countOk; 
    }
    else {
       ++$countError;
    }
}
echo $countOk.' compare ok, '.$countError.' errors';
Result: 50405 compare ok, 0 errors
 answered Jul 13, 2021 at 16:15
 

jspitjspit
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I took the visual approach (like how we do it in the real world). Instead of using formulas or what not, I solved it (or at least I think I did) by visualizing a literal calendar and then putting the dates in a multidimensional array. The first dimension corresponds to the week.
I hope someone can check if it stands your
tests. Or help someone out with a different approach.
# date in this format 2021-08-03
# week_start is either Sunday or Monday
function getWeekOfMonth($date, $week_start = "Sunday"){

    list($year, $month, $day) = explode("-", $date);

    $dates = array();
    $current_week = 1;

    $new_week_signal = $week_start == "Sunday" ? 6 : 0;

    for($i = 1; $i <= date("t", strtotime($date)); $i++){
        $current_date = strtotime("{$year}-{$month}-".$i);
        $dates[$current_week][] = $i;
        if(date('w', $current_date) == $new_week_signal){
            $current_week++;
        }
    }

    foreach($dates as $week => $days){
        if(in_array(intval($day), $days)){
            return $week;
       }
    }

    return false;

}
 answered Aug 3, 2021 at 16:48
 
PaoloPaolo
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2
//It's easy, no need to use php function
//Let's say your date is 2017-07-02

$Date = explode("-","2017-07-02");
$DateNo = $Date[2];
$WeekNo = $DateNo / 7; // devide it with 7
if(is_float($WeekNo) == true)
{
   $WeekNo = ceil($WeekNo); //So answer will be 1
}  

//If value is not float then ,you got your answer directly
 
Anders
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 answered Jul 2, 2017 at 13:44
 
1
 Not the answer you're looking for? Browse other questions tagged php arrays
date week-number or ask your own question. 
 

How to Find week of month in PHP?
Date = '2018-08-08' => Y-m-d.
 
Find out day of the month eg. ... .
 
Find out Numeric representation of the day of the week minus 1 (number of days in week) eg. ( 3-1).
 
Take difference and store in result..
 
Subtract 1 from result..
 
Divide it by 7 to result and ceil the value of result..
 
Add 1 to result eg..
 
How do I calculate the number of weeks between two dates in PHP?
php function week_between_two_dates($date1, $date2) { $first = DateTime::createFromFormat('m/d/Y', $date1); $second = DateTime::createFromFormat('m/d/Y', $date2); if($date1 > $date2) return week_between_two_dates($date2, $date1); return floor($first->diff($second)->days/7); } $dt1 = '1/1/2014'; $dt2 = '12/31/2014'; ...