I have to create a scheduling component that will plan e-mails that need to be sent out. Users can select a start time, end time, and frequency. Code should produce a random moment for every frequency, between start and end time. Outside of office hours.
Paramaters:
User can select a period between 01/01/2020 (the start) and 01/01/2021 (the end). In this case user selects a timespan of one exactly year.
User can select a frequency. In this case user selects '2 months'.
Function:
Code produces a list of datetimes. The total time (one year) is divided by frequency (2 months). We expect a list of 6 datetimes.
Every datetime is a random moment in said frequency (2 months). Within office hours.
Result:
An example result for these paramaters might as follows, with the calculated frequency bounds for clarity:
[jan/feb] 21-02-2020 11.36
[mrt/apr] 04-03-2020 16.11
[mei/jun] 13-05-2020 09.49
[jul-aug] 14-07-2020 15.25
[sep-okt] 02-09-2020 14.09
[nov-dec] 25-12-2020 13.55
--
I've been thinking about how to implement this best, but I can't figure out an elegant solution.
How could one do this using PHP?
Any insights, references, or code spikes would be greatly appreciated. I'm really stuck on this one.
I think you're just asking for suggestions on how to generate a list of repeating (2 weekly) dates with a random time between say 9am and 5pm? Is that right?
If so - something like this (untested, pseudo code) might be a starting point:
$start = new Datetime('1st January 2021');
$end = new Datetime('1st July 2021');
$day_start = 9;
$day_end = 17;
$date = $start;
$dates = [$date]; // Start date into array
while($date < $end) {
$new_date = clone($date->modify("+ 2 weeks"));
$new_date->setTime(mt_rand($day_start, $day_end), mt_rand(0, 59));
$dates[] = $new_date;
}
var_dump($dates);
Steve's anwser seems good, but you should consider 2 additional things
holiday check, in the while after first $new_date line, like:
$holiday = array('2021-01-01', '2021-01-06', '2021-12-25');
if (!in_array($new_date,$holiday))
also a check if date is a office day or a weekend in a similar way as above with working days as an array.
It's kind of crappy code but I think it will work as you wish.
function getDiffInSeconds(\DateTime $start, \DateTime $end) : int
{
$startTimestamp = $start->getTimestamp();
$endTimestamp = $end->getTimestamp();
return $endTimestamp - $startTimestamp;
}
function getShiftData(\DateTime $start, \DateTime $end) : array
{
$shiftStartHour = \DateTime::createFromFormat('H:i:s', $start->format('H:i:s'));
$shiftEndHour = \DateTime::createFromFormat('H:i:s', $end->format('H:i:s'));
$shiftInSeconds = intval($shiftEndHour->getTimestamp() - $shiftStartHour->getTimestamp());
return [
$shiftStartHour,
$shiftEndHour,
$shiftInSeconds,
];
}
function dayIsWeekendOrHoliday(\DateTime $date, array $holidays = []) : bool
{
$weekendDayIndexes = [
0 => 'Sunday',
6 => 'Saturday',
];
$dayOfWeek = $date->format('w');
if (empty($holidays)) {
$dayIsWeekendOrHoliday = isset($weekendDayIndexes[$dayOfWeek]);
} else {
$dayMonthDate = $date->format('d/m');
$dayMonthYearDate = $date->format('d/m/Y');
$dayIsWeekendOrHoliday = (isset($weekendDayIndexes[$dayOfWeek]) || isset($holidays[$dayMonthDate]) || isset($holidays[$dayMonthYearDate]));
}
return $dayIsWeekendOrHoliday;
}
function getScheduleDates(\DateTime $start, \DateTime $end, int $frequencyInSeconds) : array
{
if ($frequencyInSeconds < (24 * 60 * 60)) {
throw new \InvalidArgumentException('Frequency must be bigger than one day');
}
$diffInSeconds = getDiffInSeconds($start, $end);
// If difference between $start and $end is bigger than two days
if ($diffInSeconds > (2 * 24 * 60 * 60)) {
// If difference is bigger than 2 days we add 1 day to start and subtract 1 day from end
$start->modify('+1 day');
$end->modify('-1 day');
// Getting new $diffInSeconds after $start and $end changes
$diffInSeconds = getDiffInSeconds($start, $end);
}
if ($frequencyInSeconds > $diffInSeconds) {
throw new \InvalidArgumentException('Frequency is bigger than difference between dates');
}
$holidays = [
'01/01' => 'New Year',
'18/04/2020' => 'Easter 1st official holiday because 19/04/2020',
'20/04/2020' => 'Easter',
'21/04/2020' => 'Easter 2nd day',
'27/04' => 'Konings',
'04/05' => '4mei',
'05/05' => '4mei',
'24/12' => 'Christmas 1st day',
'25/12' => 'Christmas 2nd day',
'26/12' => 'Christmas 3nd day',
'27/12' => 'Christmas 3rd day',
'31/12' => 'Old Year'
];
[$shiftStartHour, $shiftEndHour, $shiftInSeconds] = getShiftData($start, $end);
$amountOfNotifications = floor($diffInSeconds / $frequencyInSeconds);
$periodInSeconds = intval($diffInSeconds / $amountOfNotifications);
$maxDaysBetweenNotifications = intval($periodInSeconds / (24 * 60 * 60));
// If $maxDaysBetweenNotifications is equals to 1 then we have to change $periodInSeconds to amount of seconds for one day
if ($maxDaysBetweenNotifications === 1) {
$periodInSeconds = (24 * 60 * 60);
}
$dates = [];
for ($i = 0; $i < $amountOfNotifications; $i++) {
$periodStart = clone $start;
$periodStart->setTimestamp($start->getTimestamp() + ($i * $periodInSeconds));
$seconds = mt_rand(0, $shiftInSeconds);
// If $maxDaysBetweenNotifications is equals to 1 then we have to check only one day without loop through the dates
if ($maxDaysBetweenNotifications === 1) {
$interval = new \DateInterval('P' . $maxDaysBetweenNotifications . 'DT' . $seconds . 'S');
$date = clone $periodStart;
$date->add($interval);
$dayIsWeekendOrHoliday = dayIsWeekendOrHoliday($date, $holidays);
} else {
// When $maxDaysBetweenNotifications we have to loop through the dates to pick them
$loopsCount = 0;
$maxLoops = 3; // Max loops before breaking and skipping the period
do {
$day = mt_rand(0, $maxDaysBetweenNotifications);
$periodStart->modify($shiftStartHour);
$interval = new \DateInterval('P' . $day . 'DT' . $seconds . 'S');
$date = clone $periodStart;
$date->add($interval);
$dayIsWeekendOrHoliday = dayIsWeekendOrHoliday($date, $holidays);
// If the day is weekend or holiday then we have to increment $loopsCount by 1 for each loop
if ($dayIsWeekendOrHoliday === true) {
$loopsCount++;
// If $loopsCount is equals to $maxLoops then we have to break the loop
if ($loopsCount === $maxLoops) {
break;
}
}
} while ($dayIsWeekendOrHoliday);
}
// Adds the date to $dates only if the day is not a weekend day and holiday
if ($dayIsWeekendOrHoliday === false) {
$dates[] = $date;
}
}
return $dates;
}
$start = new \DateTime('2020-12-30 08:00:00', new \DateTimeZone('Europe/Sofia'));
$end = new \DateTime('2021-01-18 17:00:00', new \DateTimeZone('Europe/Sofia'));
$frequencyInSeconds = 86400; // 1 day
$dates = getScheduleDates($start, $end, $frequencyInSeconds);
var_dump($dates);
You have to pass $start, $end and $frequencyInSeconds as I showed in example and then you will get your random dates. Notice that I $start and $end must have hours in them because they are used as start and end hours for shifts. Because the rule is to return a date within a shift time only in working days. Also you have to provide frequency in seconds - you can calculate them outside the function or you can change it to calculate them inside. I did it this way because I don't know what are your predefined periods.
This function returns an array of \DateTime() instances so you can do whatever you want with them.
UPDATE 08/01/2020:
Holidays now are part of calculation and they will be excluded from returned dates if they are passed when you are calling the function. You can pass them in d/m and d/m/Y formats because of holidays like Easter and in case when the holiday is on weekend but people will get additional dayoff during the working week.
UPDATE 13/01/2020:
I've made updated code version to fix the issue with infinite loops when $frequencyInSeconds is shorter like 1 day. The new code used few functions getDiffInSeconds, getShiftData and dayIsWeekendOrHoliday as helper methods to reduce code duplication and cleaner and more readable code
I have a function to return the difference between 2 dates, however I need to work out the difference in working hours, assuming Monday to Friday (9am to 5:30pm):
//DATE DIFF FUNCTION
// Set timezone
date_default_timezone_set("GMT");
// Time format is UNIX timestamp or
// PHP strtotime compatible strings
function dateDiff($time1, $time2, $precision = 6) {
// If not numeric then convert texts to unix timestamps
if (!is_int($time1)) {
$time1 = strtotime($time1);
}
if (!is_int($time2)) {
$time2 = strtotime($time2);
}
// If time1 is bigger than time2
// Then swap time1 and time2
if ($time1 > $time2) {
$ttime = $time1;
$time1 = $time2;
$time2 = $ttime;
}
// Set up intervals and diffs arrays
$intervals = array('year','month','day','hour','minute','second');
$diffs = array();
// Loop thru all intervals
foreach ($intervals as $interval) {
// Set default diff to 0
$diffs[$interval] = 0;
// Create temp time from time1 and interval
$ttime = strtotime("+1 " . $interval, $time1);
// Loop until temp time is smaller than time2
while ($time2 >= $ttime) {
$time1 = $ttime;
$diffs[$interval]++;
// Create new temp time from time1 and interval
$ttime = strtotime("+1 " . $interval, $time1);
}
}
$count = 0;
$times = array();
// Loop thru all diffs
foreach ($diffs as $interval => $value) {
// Break if we have needed precission
if ($count >= $precision) {
break;
}
// Add value and interval
// if value is bigger than 0
if ($value > 0) {
// Add s if value is not 1
if ($value != 1) {
$interval .= "s";
}
// Add value and interval to times array
$times[] = $value . " " . $interval;
$count++;
}
}
// Return string with times
return implode(", ", $times);
}
Date 1 = 2012-03-24 03:58:58
Date 2 = 2012-03-22 11:29:16
Is there a simple way of doing this, i.e - calculating the percentage of working hours in a week and dividing the difference using the above function - I have played around with this idea and got some very strange figures...
Or is there better way....?
This example uses PHP's built in DateTime classes to do the date math. How I approached this was to start by counting the number of full working days between the two dates and then multiply that by 8 (see notes). Then it gets the hours worked on the partial days and adds them to the total hours worked. Turning this into a function would be fairly straightforward to do.
Notes:
Does not take timestamps into account. But you already know how to do that.
Does not handle holidays. (That can be easily added by using an array of holidays and adding it to where you filter out Saturdays and Sundays).
Requires PHP 5.3.6+
Assumes an 8 hour workday. If employees do not take lunch change $hours = $days * 8; to $hours = $days * 8.5;
.
<?php
// Initial datetimes
$date1 = new DateTime('2012-03-22 11:29:16');
$date2 = new DateTime('2012-03-24 03:58:58');
// Set first datetime to midnight of next day
$start = clone $date1;
$start->modify('+1 day');
$start->modify('midnight');
// Set second datetime to midnight of that day
$end = clone $date2;
$end->modify('midnight');
// Count the number of full days between both dates
$days = 0;
// Loop through each day between two dates
$interval = new DateInterval('P1D');
$period = new DatePeriod($start, $interval, $end);
foreach ($period as $dt) {
// If it is a weekend don't count it
if (!in_array($dt->format('l'), array('Saturday', 'Sunday'))) {
$days++;
}
}
// Assume 8 hour workdays
$hours = $days * 8;
// Get the number of hours worked on the first day
$date1->modify('5:30 PM');
$diff = $date1->diff($start);
$hours += $diff->h;
// Get the number of hours worked the second day
$date1->modify('8 AM');
$diff = $date2->diff($end);
$hours += $diff->h;
echo $hours;
See it in action
Reference
DateTime Class
DatePeriod Class
DateInterval Class
Here's what I've come up with.
My solution checks the start and end times of the original dates, and adjusts them according to the actual start and end times of the work day (if the original start time is before work's opening time, it sets it to the latter).
After this is done to both start and end times, the times are compared to retrieve a DateInterval diff, calculating the total days, hours, etc. The date range is then checked for any weekend days, and if found, one total day is reduced from the diff.
Finally, the hours are calculated as commented. :)
Cheers to John for inspiring some of this solution, particularly the DatePeriod to check for weekends.
Gold star to anyone who breaks this; I'll be happy to update if anyone finds a loophole!
Gold star to myself, I broke it! Yeah, weekends are still buggy (try starting at 4pm on Saturday and ending at 1pm Monday). I will conquer you, work hours problem!
Ninja edit #2: I think I took care of the weekend bugs by reverting the start and end times to the most recent respective weekday if they fall on a weekend. Got good results after testing a handful of date ranges (starting and ending on the same weekend barfs, as expected). I'm not entirely convinced this is as optimized / simple as it could be, but at least it works better now.
// Settings
$workStartHour = 9;
$workStartMin = 0;
$workEndHour = 17;
$workEndMin = 30;
$workdayHours = 8.5;
$weekends = ['Saturday', 'Sunday'];
$hours = 0;
// Original start and end times, and their clones that we'll modify.
$originalStart = new DateTime('2012-03-22 11:29:16');
$start = clone $originalStart;
// Starting on a weekend? Skip to a weekday.
while (in_array($start->format('l'), $weekends))
{
$start->modify('midnight tomorrow');
}
$originalEnd = new DateTime('2012-03-24 03:58:58');
$end = clone $originalEnd;
// Ending on a weekend? Go back to a weekday.
while (in_array($end->format('l'), $weekends))
{
$end->modify('-1 day')->setTime(23, 59);
}
// Is the start date after the end date? Might happen if start and end
// are on the same weekend (whoops).
if ($start > $end) throw new Exception('Start date is AFTER end date!');
// Are the times outside of normal work hours? If so, adjust.
$startAdj = clone $start;
if ($start < $startAdj->setTime($workStartHour, $workStartMin))
{
// Start is earlier; adjust to real start time.
$start = $startAdj;
}
else if ($start > $startAdj->setTime($workEndHour, $workEndMin))
{
// Start is after close of that day, move to tomorrow.
$start = $startAdj->setTime($workStartHour, $workStartMin)->modify('+1 day');
}
$endAdj = clone $end;
if ($end > $endAdj->setTime($workEndHour, $workEndMin))
{
// End is after; adjust to real end time.
$end = $endAdj;
}
else if ($end < $endAdj->setTime($workStartHour, $workStartMin))
{
// End is before start of that day, move to day before.
$end = $endAdj->setTime($workEndHour, $workEndMin)->modify('-1 day');
}
// Calculate the difference between our modified days.
$diff = $start->diff($end);
// Go through each day using the original values, so we can check for weekends.
$period = new DatePeriod($start, new DateInterval('P1D'), $end);
foreach ($period as $day)
{
// If it's a weekend day, take it out of our total days in the diff.
if (in_array($day->format('l'), ['Saturday', 'Sunday'])) $diff->d--;
}
// Calculate! Days * Hours in a day + hours + minutes converted to hours.
$hours = ($diff->d * $workdayHours) + $diff->h + round($diff->i / 60, 2);
As the old saying goes "if you want something done right do it yourself". Not saying this is optimal but its atleast returning the correct amount of hours for me.
function biss_hours($start, $end){
$startDate = new DateTime($start);
$endDate = new DateTime($end);
$periodInterval = new DateInterval( "PT1H" );
$period = new DatePeriod( $startDate, $periodInterval, $endDate );
$count = 0;
foreach($period as $date){
$startofday = clone $date;
$startofday->setTime(8,30);
$endofday = clone $date;
$endofday->setTime(17,30);
if($date > $startofday && $date <= $endofday && !in_array($date->format('l'), array('Sunday','Saturday'))){
$count++;
}
}
//Get seconds of Start time
$start_d = date("Y-m-d H:00:00", strtotime($start));
$start_d_seconds = strtotime($start_d);
$start_t_seconds = strtotime($start);
$start_seconds = $start_t_seconds - $start_d_seconds;
//Get seconds of End time
$end_d = date("Y-m-d H:00:00", strtotime($end));
$end_d_seconds = strtotime($end_d);
$end_t_seconds = strtotime($end);
$end_seconds = $end_t_seconds - $end_d_seconds;
$diff = $end_seconds-$start_seconds;
if($diff!=0):
$count--;
endif;
$total_min_sec = date('i:s',$diff);
return $count .":".$total_min_sec;
}
$start = '2014-06-23 12:30:00';
$end = '2014-06-27 15:45:00';
$go = biss_hours($start,$end);
echo $go;
I know about the unwanted behaviour of PHP's function
strtotime
For example, when adding a month (+1 month) to dates like: 31.01.2011 -> 03.03.2011
I know it's not officially a PHP bug, and that this solution has some arguments behind it, but at least for me, this behavior has caused a lot waste of time (in the past and present) and I personally hate it.
What I found even stranger is that for example in:
MySQL: DATE_ADD('2011-01-31', INTERVAL 1 MONTH) returns 2011-02-28
or
C# where new DateTime(2011, 01, 31).AddMonths(1); will return 28.02.2011
wolframalpha.com giving 31.01.2013 + 1 month as input; will return Thursday, February 28, 2013
It sees to me that others have found a more decent solution to the stupid question that I saw alot in PHP bug reports "what day will it be, if I say we meet in a month from now" or something like that. The answer is: if 31 does not exists in next month, get me the last day of that month, but please stick to next month.
So MY QUESTION IS: is there a PHP function (written by somebody) that resolves this not officially recognized bug? As I don't think I am the only one who wants another behavior when adding / subtracting months.
I am particulary interested in solutions what also work not just for the end of the month, but a complete replacement of strtotime. Also the case strotime +n months should be also dealt with.
Happy coding!
what you need is to tell PHP to be smarter
$the_date = strtotime('31.01.2011');
echo date('r', strtotime('last day of next month', $the_date));
$the_date = strtotime('31.03.2011');
echo date('r', strtotime('last day of next month', $the_date));
assuming you are only interesting on the last day of next month
reference - http://www.php.net/manual/en/datetime.formats.relative.php
PHP devs surely don't consider this as bug. But in strtotime's docs there are few comments with solutions for your problem (look for 28th Feb examples ;)), i.e. this one extending DateTime class:
<?php
// this will give us 2010-02-28 ()
echo PHPDateTime::DateNextMonth(strftime('%F', strtotime("2010-01-31 00:00:00")), 31);
?>
Class PHPDateTime:
<?php
/**
* IA FrameWork
* #package: Classes & Object Oriented Programming
* #subpackage: Date & Time Manipulation
* #author: ItsAsh <ash at itsash dot co dot uk>
*/
final class PHPDateTime extends DateTime {
// Public Methods
// ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
/**
* Calculate time difference between two dates
* ...
*/
public static function TimeDifference($date1, $date2)
$date1 = is_int($date1) ? $date1 : strtotime($date1);
$date2 = is_int($date2) ? $date2 : strtotime($date2);
if (($date1 !== false) && ($date2 !== false)) {
if ($date2 >= $date1) {
$diff = ($date2 - $date1);
if ($days = intval((floor($diff / 86400))))
$diff %= 86400;
if ($hours = intval((floor($diff / 3600))))
$diff %= 3600;
if ($minutes = intval((floor($diff / 60))))
$diff %= 60;
return array($days, $hours, $minutes, intval($diff));
}
}
return false;
}
/**
* Formatted time difference between two dates
*
* ...
*/
public static function StringTimeDifference($date1, $date2) {
$i = array();
list($d, $h, $m, $s) = (array) self::TimeDifference($date1, $date2);
if ($d > 0)
$i[] = sprintf('%d Days', $d);
if ($h > 0)
$i[] = sprintf('%d Hours', $h);
if (($d == 0) && ($m > 0))
$i[] = sprintf('%d Minutes', $m);
if (($h == 0) && ($s > 0))
$i[] = sprintf('%d Seconds', $s);
return count($i) ? implode(' ', $i) : 'Just Now';
}
/**
* Calculate the date next month
*
* ...
*/
public static function DateNextMonth($now, $date = 0) {
$mdate = array(0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31);
list($y, $m, $d) = explode('-', (is_int($now) ? strftime('%F', $now) : $now));
if ($date)
$d = $date;
if (++$m == 2)
$d = (($y % 4) === 0) ? (($d <= 29) ? $d : 29) : (($d <= 28) ? $d : 28);
else
$d = ($d <= $mdate[$m]) ? $d : $mdate[$m];
return strftime('%F', mktime(0, 0, 0, $m, $d, $y));
}
}
?>
Here's the algorithm you can use. It should be simple enough to implement yourself.
Have the original date and the +1 month date in variables
Extract the month part of both variables
If the difference is greater than 1 month (or if the original is December and the other is not January) change the latter variable to the last day of the next month. You can use for example t in date() to get the last day: date( 't.m.Y' )
Had the same issue recently and ended up writing a class that handles adding/subtracting various time intervals to DateTime objects.
Here's the code:
https://gist.github.com/pavlepredic/6220041#file-gistfile1-php
I've been using this class for a while and it seems to work fine, but I'm really interested in some peer review. What you do is create a TimeInterval object (in your case, you would specify 1 month as the interval) and then call addToDate() method, making sure you set $preventMonthOverflow argument to true. The code will make sure that the resulting date does not overflow into next month.
Sample usage:
$int = new TimeInterval(1, TimeInterval::MONTH);
$date = date_create('2013-01-31');
$future = $int->addToDate($date, true);
echo $future->format('Y-m-d');
Resulting date is:
2013-02-28
Here is an implementation of an improved version of Juhana's answer above:
<?php
function sameDateNextMonth(DateTime $createdDate, DateTime $currentDate) {
$addMon = clone $currentDate;
$addMon->add(new DateInterval("P1M"));
$nextMon = clone $currentDate;
$nextMon->modify("last day of next month");
if ($addMon->format("n") == $nextMon->format("n")) {
$recurDay = $createdDate->format("j");
$daysInMon = $addMon->format("t");
$currentDay = $currentDate->format("j");
if ($recurDay > $currentDay && $recurDay <= $daysInMon) {
$addMon->setDate($addMon->format("Y"), $addMon->format("n"), $recurDay);
}
return $addMon;
} else {
return $nextMon;
}
}
This version takes $createdDate under the presumption that you are dealing with a recurring monthly period, such as a subscription, that started on a specific date, such as the 31st. It always takes $createdDate so late "recurs on" dates won't shift to lower values as they are pushed forward thru lesser-valued months (e.g., so all 29th, 30th or 31st recur dates won't eventually get stuck on the 28th after passing thru a non-leap-year February).
Here is some driver code to test the algorithm:
$createdDate = new DateTime("2015-03-31");
echo "created date = " . $createdDate->format("Y-m-d") . PHP_EOL;
$next = sameDateNextMonth($createdDate, $createdDate);
echo " next date = " . $next->format("Y-m-d") . PHP_EOL;
foreach(range(1, 12) as $i) {
$next = sameDateNextMonth($createdDate, $next);
echo " next date = " . $next->format("Y-m-d") . PHP_EOL;
}
Which outputs:
created date = 2015-03-31
next date = 2015-04-30
next date = 2015-05-31
next date = 2015-06-30
next date = 2015-07-31
next date = 2015-08-31
next date = 2015-09-30
next date = 2015-10-31
next date = 2015-11-30
next date = 2015-12-31
next date = 2016-01-31
next date = 2016-02-29
next date = 2016-03-31
next date = 2016-04-30
I have solved it by this way:
$startDate = date("Y-m-d");
$month = date("m",strtotime($startDate));
$nextmonth = date("m",strtotime("$startDate +1 month"));
if((($nextmonth-$month) > 1) || ($month == 12 && $nextmonth != 1))
{
$nextDate = date( 't.m.Y',strtotime("$initialDate +1 week"));
}else
{
$nextDate = date("Y-m-d",strtotime("$initialDate +1 month"));
}
echo $nextDate;
Somewhat similar to the Juhana's answer but more intuitive and less complications expected. Idea is like this:
Store original date and the +n month(s) date in variables
Extract the day part of both variables
If days do not match, subtract number of days from the future date
Plus side of this solution is that works for any date (not just the border dates) and it also works for subtracting months (by putting - instead of +).
Here is an example implementation:
$start = mktime(0,0,0,1,31,2015);
for ($contract = 0; $contract < 12; $contract++) {
$end = strtotime('+ ' . $contract . ' months', $start);
if (date('d', $start) != date('d', $end)) {
$end = strtotime('- ' . date('d', $end) . ' days', $end);
}
echo date('d-m-Y', $end) . '|';
}
And the output is following:
31-01-2015|28-02-2015|31-03-2015|30-04-2015|31-05-2015|30-06-2015|31-07-2015|31-08-2015|30-09-2015|31-10-2015|30-11-2015|31-12-2015|
function ldom($m,$y){
//return tha last date of a given month based on the month and the year
//(factors in leap years)
$first_day= strtotime (date($m.'/1/'.$y));
$next_month = date('m',strtotime ( '+32 day' , $first_day)) ;
$last_day= strtotime ( '-1 day' , strtotime (date($next_month.'/1/'.$y)) ) ;
return $last_day;
}
So I have a script that returns the number of weeks in a particular month and year. How can I take a specific day from that month and determine if it is part of week 1,2,3,4 or 5 of that month?
The most frustrating thing I have ever tried to get working - but here it is!
<?php
/**
* Returns the amount of weeks into the month a date is
* #param $date a YYYY-MM-DD formatted date
* #param $rollover The day on which the week rolls over
*/
function getWeeks($date, $rollover)
{
$cut = substr($date, 0, 8);
$daylen = 86400;
$timestamp = strtotime($date);
$first = strtotime($cut . "00");
$elapsed = ($timestamp - $first) / $daylen;
$weeks = 1;
for ($i = 1; $i <= $elapsed; $i++)
{
$dayfind = $cut . (strlen($i) < 2 ? '0' . $i : $i);
$daytimestamp = strtotime($dayfind);
$day = strtolower(date("l", $daytimestamp));
if($day == strtolower($rollover)) $weeks ++;
}
return $weeks;
}
//
echo getWeeks("2011-06-11", "sunday"); //outputs 2, for the second week of the month
?>
Edit: so much for "single line" - needed variables to avoid recomputation with the conditional. Tossed in a default argument while I was at it.
function weekOfMonth($when = null) {
if ($when === null) $when = time();
$week = date('W', $when); // note that ISO weeks start on Monday
$firstWeekOfMonth = date('W', strtotime(date('Y-m-01', $when)));
return 1 + ($week < $firstWeekOfMonth ? $week : $week - $firstWeekOfMonth);
}
Please note that weekOfMonth(strtotime('Oct 31, 2011')); will return 6; some rare months have 6 weeks in them, contrary to OP's expectation. January 2017 is another month with 6 ISO weeks - Sunday the 1st falls in the last year's week, since ISO weeks start on Monday.
For starshine531, to return a 0 indexed week of the month, change the return 1 + to return 0 + or return (int).
For Justin Stayton, for weeks starting on Sunday instead of Monday I would use strftime('%U' instead of date('W', as follows:
function weekOfMonth($when = null) {
if ($when === null) $when = time();
$week = strftime('%U', $when); // weeks start on Sunday
$firstWeekOfMonth = strftime('%U', strtotime(date('Y-m-01', $when)));
return 1 + ($week < $firstWeekOfMonth ? $week : $week - $firstWeekOfMonth);
}
For this version, 2017-04-30 is now in week 6 of April, while 2017-01-31 is now in week 5.
public function getWeeks($timestamp)
{
$maxday = date("t",$timestamp);
$thismonth = getdate($timestamp);
$timeStamp = mktime(0,0,0,$thismonth['mon'],1,$thismonth['year']); //Create time stamp of the first day from the give date.
$startday = date('w',$timeStamp); //get first day of the given month
$day = $thismonth['mday'];
$weeks = 0;
$week_num = 0;
for ($i=0; $i<($maxday+$startday); $i++) {
if(($i % 7) == 0){
$weeks++;
}
if($day == ($i - $startday + 1)){
$week_num = $weeks;
}
}
return $week_num;
}
Hello all i have been struggling for the whole day trying to figure this code out, i finally figured it out so i thought i would share it with you all.
all you need to do is put a time stamp into the function and it will return the week number back to you.
thanks
there is a problem with this method. if the passing date (Lets say 2012/01/01 which is a Sunday) and "$rollover" day is "Sunday", then this function will return 2. where its actually is 1'st week. i think i have fixed it in following function.
please add comments to make it better.
function getWeeks($date, $rollover)
{
$cut = substr($date, 0, 8);
$daylen = 86400;
$timestamp = strtotime($date);
$first = strtotime($cut . "01");
$elapsed = (($timestamp - $first) / $daylen)+1;
$i = 1;
$weeks = 0;
for($i==1; $i<=$elapsed; $i++)
{
$dayfind = $cut . (strlen($i) < 2 ? '0' . $i : $i);
$daytimestamp = strtotime($dayfind);
$day = strtolower(date("l", $daytimestamp));
if($day == strtolower($rollover))
{
$weeks++;
}
}
if($weeks==0)
{
$weeks++;
}
return $weeks;
}
This is a solution based on sberry's mathematical solution but using the PHP DateTime class instead.
function week_of_month($date) {
$first_of_month = new DateObject($date->format('Y/m/1'));
$day_of_first = $first_of_month->format('N');
$day_of_month = $date->format('j');
return floor(($day_of_first + $day_of_month - 1) / 7) + 1;
}
Just Copy and Past the code and pass month and year.
e.g month=04 year=2013.
That's exactly what You Need.
$mm= $_REQUEST['month'];
$yy= $_REQUEST['year'];
$startdate=date($yy."-".$mm."-01") ;
$current_date=date('Y-m-t');
$ld= cal_days_in_month(CAL_GREGORIAN, $mm, $yy);
$lastday=$yy.'-'.$mm.'-'.$ld;
$start_date = date('Y-m-d', strtotime($startdate));
$end_date = date('Y-m-d', strtotime($lastday));
$end_date1 = date('Y-m-d', strtotime($lastday." + 6 days"));
$count_week=0;
$week_array = array();
for($date = $start_date; $date <= $end_date1; $date = date('Y-m-d', strtotime($date. ' + 7 days')))
{
$getarray=getWeekDates($date, $start_date, $end_date);
echo "<br>";
$week_array[]=$getarray;
echo "\n";
$count_week++;
}
// its give the number of week for the given month and year
echo $count_week;
//print_r($week_array);
function getWeekDates($date, $start_date, $end_date)
{
$week = date('W', strtotime($date));
$year = date('Y', strtotime($date));
$from = date("Y-m-d", strtotime("{$year}-W{$week}+1"));
if($from < $start_date) $from = $start_date;
$to = date("Y-m-d", strtotime("{$year}-W{$week}-6"));
if($to > $end_date) $to = $end_date;
$array1 = array(
"ssdate" => $from,
"eedate" => $to,
);
return $array1;
// echo "Start Date-->".$from."End Date -->".$to;
}
for($i=0;$i<$count_week;$i++)
{
$start= $week_array[$i]['ssdate'];
echo "--";
$week_array[$i]['eedate'];
echo "<br>";
}
OUTPUT:
week( 0 )=>2013-03-01---2013-03-02
week( 1 )=>2013-03-03---2013-03-09
week( 2 )=>2013-03-10---2013-03-16
week( 3 )=>2013-03-17---2013-03-23
week( 4 )=>2013-03-24---2013-03-30
week( 5 )=>2013-03-31---2013-03-31
I think I found an elegant solution
$time = time(); // or whenever
$week_of_the_month = ceil(date('d', $time)/7);
For a Monday-Sunday (ISO 8601) week (or, if you simply don't care), you can do this in one line:
function get_week_of_month($date) {
return date('W', $date) - date('W', strtotime(date("Y-m-01", $date))) + 1;
}
(Source)
For anything else, (e.g. a Sunday-Saturday week), you just need to tweak $date inside the function:
function get_week_of_month($date) {
$date += 86400; //For weeks starting on Sunday
return date('W', $date) - date('W', strtotime(date("Y-m-01", $date))) + 1;
}
(Thanks to these guys/gals)
NOTE: You may run into some issues at the end of the year (e.g. around 12/31, 1/1, etc.). Read more here.
This is the snippet that I made to fulfill my requirements for the same. Hope this will help you.
function getWeek($timestamp) {
$week_year = date('W',$timestamp);
$week = 0;//date('d',$timestamp)/7;
$year = date('Y',$timestamp);
$month = date('m',$timestamp);
$day = date('d',$timestamp);
$prev_month = date('m',$timestamp) -1;
if($month != 1 ){
$last_day_prev = $year."-".$prev_month."-1";
$last_day_prev = date('t',strtotime($last_day_prev));
$week_year_last_mon = date('W',strtotime($year."-".$prev_month."-".$last_day_prev));
$week_year_first_this = date('W',strtotime($year."-".$month."-1"));
if($week_year_first_this == $week_year_last_mon){
$week_diff = 0;
}
else{
$week_diff = 1;
}
if($week_year ==1 && $month == 12 ){
// to handle December's last two days coming in first week of January
$week_year = 53;
}
$week = $week_year-$week_year_last_mon + 1 +$week_diff;
}
else{
// to handle first three days January coming in last week of December.
$week_year_first_this = date('W',strtotime($year."-01-1"));
if($week_year_first_this ==52 || $week_year_first_this ==53){
if($week_year == 52 || $week_year == 53){
$week =1;
}
else{
$week = $week_year + 1;
}
}
else{
$week = $week_year;
}
}
return $week;
}
This is probably not a good way to do this but it's my first thought and I'm really tired.
Put all your dates into an array. The date object must have a day name (Monday). Create a method that searches the array and when ever you hit a Sunday you add 1 to a week counter. Once you find the date you're looking for return the week counter. That is the week the day falls in of the year. For the week in the month you have to reset the week counter every time you get to the last day in each month.
Here comes two liner:
function getWeekOfMonth(DateTime $date) {
$firstDayOfMonth = new DateTime($date->format('Y-m-1'));
return ceil(($firstDayOfMonth->format('N') + $date->format('j') - 1) / 7);
}
And Wtower's solutions doesn't work 100% properly.
Thought I'd share my function as well. This returns an array of weeks. Every week is an array with weeks day (0..6) as key and months day (1..31) as value.
Function assumes that week starts with Sunday.
Enjoy!
function get_weeks($year, $month){
$days_in_month = date("t", mktime(0, 0, 0, $month, 1, $year));
$weeks_in_month = 1;
$weeks = array();
//loop through month
for ($day=1; $day<=$days_in_month; $day++) {
$week_day = date("w", mktime(0, 0, 0, $month, $day, $year));//0..6 starting sunday
$weeks[$weeks_in_month][$week_day] = $day;
if ($week_day == 6) {
$weeks_in_month++;
}
}
return $weeks;
}
My 5 cents:
/**
* calculate number of weeks in a particular month
*/
function weeksInMonth($month=null,$year=null){
if( null==($year) ) {
$year = date("Y",time());
}
if(null==($month)) {
$month = date("m",time());
}
// find number of days in this month
$daysInMonths = date('t',strtotime($year.'-'.$month.'-01'));
$numOfweeks = ($daysInMonths%7==0?0:1) + intval($daysInMonths/7);
$monthEndingDay= date('N',strtotime($year.'-'.$month.'-'.$daysInMonths));
$monthStartDay = date('N',strtotime($year.'-'.$month.'-01'));
if($monthEndingDay<$monthStartDay){
$numOfweeks++;
}
return $numOfweeks;
}
I create this function, from brazil :) I hope it is useful
function weekofmonth($time) {
$firstday = 1;
$lastday = date('j',$time);
$lastdayweek = 6; //Saturday
$week = 1;
for ($day=1;$day<=$lastday;$day++) {
$timetmp = mktime(0, 0, 0, date('n',$time), $day, date('Y',$time));
if (date('N',$timetmp) == $lastdayweek) {
$week++;
}
}
if (date('N',$time)==$lastdayweek) {
$week--;
}
return $week;
}
$time = mktime(0, 0, 0, 9, 30, 2014);
echo weekofmonth($time);
I found a easy way to determine what week of the month today is in, and it would be a small change to have it work on any other date. I'm adding my two cents in here as I think my way is much more compact then the methods listed.
$monthstart = date("N",strtotime(date("n/1/Y")));
$date =( date("j")+$monthstart ) /7;
$ddate= floor( $date );
if($ddate != date) {$ddate++;}
and $ddate contains the week number you could modify it like so
function findweek($indate)
{
$monthstart = date("N",strtotime(date("n/1/Y",strtotime($indate))));
$date =( date("j",strtotime($indate))+$monthstart ) /7;
$ddate= floor( $date );
if($ddate != $date) {$ddate++;}
return $ddate;
}
and it would return what week of the month any date you give it is.
what it does is first find the number of days from the start of the week to the first of the month. then adds that on to the current date then divides the new date by 7 and that will give you how many weeks have passed since the start of the month, including a decimal place for the part of the the current week that has passed. so what I do next is round down that number, then compare the rounded down version to the original if the two match your at the end of the week so it's already in the number. if they don't then just add one to the rounded down number and voila you have the current week number.
Srahul07's solution works perfectly... If you abide by the Monday-Sunday week system! Here in 'murica, non-business folk tend to go by Sunday-Saturday being a week, so May 1, 2011 is week 1 and May 2, 2011 is still week 1.
Adding the following logic to the bottom of his function, right before it returns $week will convert this to a Sunday -> Monday system:
if (!date('w',strtotime("$year-$month-01")) && date('w',$timestamp))
$week--;
elseif (date('w',strtotime("$year-$month-01")) && !date('w',$timestamp))
$week++;
After alot of efoort i found the solution
<?php
function getWeeks($month,$year)
{
$month = intval($month); //force month to single integer if '0x'
$suff = array('st','nd','rd','th','th','th'); //week suffixes
$end = date('t',mktime(0,0,0,$month,1,$year)); //last date day of month: 28 - 31
$start = date('w',mktime(0,0,0,$month,1,$year)); //1st day of month: 0 - 6 (Sun - Sat)
$last = 7 - $start; //get last day date (Sat) of first week
$noweeks = ceil((($end - ($last + 1))/7) + 1); //total no. weeks in month
$output = ""; //initialize string
$monthlabel = str_pad($month, 2, '0', STR_PAD_LEFT);
for($x=1;$x<$noweeks+1;$x++)
{
if($x == 1)
{
$startdate = "$year-$monthlabel-01";
$day = $last - 6;
}
else
{
$day = $last + 1 + (($x-2)*7);
$day = str_pad($day, 2, '0', STR_PAD_LEFT);
$startdate = "$year-$monthlabel-$day";
}
if($x == $noweeks)
{
$enddate = "$year-$monthlabel-$end";
}
else
{
$dayend = $day + 6;
$dayend = str_pad($dayend, 2, '0', STR_PAD_LEFT);
$enddate = "$year-$monthlabel-$dayend";
}
$j=1;
if($j--)
{
$k=getTotalDate($startdate,$enddate);
$j=1;
}
$output .= "Week ".$xyz." week -> Start date=$startdate End date=$enddate <br />";
}
return $output;
}
if(isset($_POST) && !empty($_POST)){
$month = $_POST['m'];
$year = $_POST['y'];
echo getWeeks($month,$year);
}
?>
<form method="post">
M:
<input name="m" value="" />
Y:
<input name="y" value="" />
<input type="submit" value="go" />
</form>
I really liked #michaelc's answer. However, I got stuck on a few points. It seemed that every time Sunday rolled around, there was an offset of one. I think it has to do with what day of the week is the start of the week. In any case, here is my slight alteration to it, expanded a bit for readability:
function wom(\DateTime $date) {
// The week of the year of the current month
$cw = date('W', $date->getTimestamp());
// The week of the year of the first of the given month
$fw = date('W',strtotime(date('Y-m-01',$date->getTimeStamp())));
// Offset
$o = 1;
// If it is a Saturday, offset by two.
if( date('N',$date->getTimestamp()) == 7 ) {
$o = 2;
}
return $cw -$fw + $o;
}
So if the date is Nov. 9, 2013...
$cw = 45
$fw = 44
and with the offset of 1, it correctly returns 2.
If the date is Nov. 10, 2013, $cw and $fw are the same as before, but the offset is 2, and it correctly returns 3.
function get_week_of_month( $timestamp )
{
$week_of_month = 0;
$month = date( 'j', $timestamp );
$test_month = $month;
while( $test_month == $month )
{
$week_of_month++;
$timestamp = strtotime( '-1 week', $timestamp );
$test_month = date( 'j', $timestamp );
}
return $week_of_month;
}
I found this online:
http://kcwebprogrammers.blogspot.de/2009/03/current-week-in-month-php.html
He has a very simple solution which seems to work fine for me.
$currentWeek = ceiling((date("d") - date("w") - 1) / 7) + 1;
So for example:
$now = strtotime("today");
$weekOfMonth = ceil((date("d", $now) - date("w", $now) - 1) / 7) + 1;
you can use W in newer php versions. http://php.net/manual/en/function.date.php
i have used it like so:
function getWeek($date) {
$month_start=strtotime("1 ".date('F Y',$date));
$current_date=strtotime(date('j F Y',$date));
$month_week=date("W",$month_start);
$current_week=date("W",$current_date);
return ($current_week-$month_week);
}//0 is the week of the first.
Short and foolproof:
// Function accepts $date as a string,
// Returns the week number in which the given date falls.
// Assumed week starts on Sunday.
function wom($date) {
$date = strtotime($date);
$weeknoofday = date('w', $date);
$day = date('j', $date);
$weekofmonth = ceil(($day + (7-($weeknoofday+1))) / 7);
return $weekofmonth;
}
// Test
foreach (range(1, 31) as $day) {
$test_date = "2015-01-" . str_pad($day, 2, '0', STR_PAD_LEFT);
echo "$test_date - ";
echo wom($test_date) . "\n";
}
I use this simple function:
function weekNumberInMonth($timestampDate)
{
$firstDayOfMonth = strtotime(date('01-M-Y 00:00:00', $timestampDate));
$firstWeekdayOfMonth = date( 'w', $firstDayOfMonth);
$dayNumberInMonth = date('d', $timestampDate);
$weekNumberInMonth = ceil(($dayNumberInMonth + $firstWeekdayOfMonth) / 7);
return $weekNumberInMonth;
}
if I understand correct, the question is how to identify what number of week within a month of a specific day... I was looking for similar solution. I used some ideas of above answers to develop my own solution. Hope it can be helpful for somebody. If Yes, then UpVote my answer.
function week_number_within_month($datenew){
$year = date("Y",strtotime($datenew));
$month = date("m",strtotime($datenew));
// find number of days in this month
$daysInMonths = date('t',strtotime($year.'-'.$month.'-01'));
$numOfweeks = ($daysInMonths%7==0?0:1) + intval($daysInMonths/7);
$monthEndingDay= date('N',strtotime($year.'-'.$month.'-'.$daysInMonths));
$monthStartDay = date('N',strtotime($year.'-'.$month.'-01'));
if($monthEndingDay<$monthStartDay){
$numOfweeks++;
}
$date=date('Y/m/d', strtotime($year.'-'. $month.'-01'));
$week_array=Array();
for ($i=1; $i<=$numOfweeks; $i++){ /// create an Array of all days of month separated by weeks as a keys
$max = 7;
if ($i ==1){ $max = 8 - $monthStartDay;}
if ($i == $numOfweeks){ $max = $monthEndingDay;}
for ($r=1; $r<=$max; $r++){
$week_array[$i][]=$date;
$date = date('Y/m/d',strtotime($date . "+1 days"));
}
}
$new_datenew = date('Y/m/d', strtotime($datenew));
$week_result='';
foreach ($week_array as $key => $val){ /// finding what week number of my date from week_array
foreach ($val as $kr => $value){
if ($new_datenew == $value){
$week_result = $key;
}
}
}
return $week_result;
}
print week_number_within_month('2016-09-15');
function getWeekOfMonth(\DateTime $date)
{
$firstWeekdayOfMonth = new DateTime("first weekday 0 {$date->format('M')} {$date->format('Y')}");
$offset = $firstWeekdayOfMonth->format('N')-1;
return intval(($date->format('j') + $offset)/7)+1;
}
/**
* In case of Week we can get the week of year. So whenever we will get the week of the month then we have to
* subtract the until last month weeks from it will give us the current month week.
*/
$dateComponents = getdate();
if($dateComponents['mon'] == 1)
$weekOfMonth = date('W', strtotime($dateComponents['year'].'-'.$dateComponents['mon'].'-'.$dateComponents['mday']))-1; // We subtract -1 to map it to the array
else
$weekOfMonth = date('W', strtotime($dateComponents['year'].'-'.$dateComponents['mon'].'-'.$dateComponents['mday']))-date('W', strtotime($dateComponents['year'].'-'.$dateComponents['mon'].'-01'));
Using Carbon:
$date = Carbon::now();
$d1 = $date->startOfMonth();
$d2 = $date->endOfMonth();
$weeks = $d1->diffInWeeks($d2);
If you clearly want to separate a month into 4 Weeks, you can use this function.
This is helpful, if you want
"the first monday of month"
"the third thursday of month" etc.
Here we go
/**
* This Calculates (and returns) the week number within a month, based on date('j') day of month.
* This is useful, if you want to have (for instance) the first Thu in month, regardless of date
* #param $Timestamp
* #return float|int
*/
function getWeekOfMonth($Timestamp)
{
$DayOfMonth=date('j', $Timestamp); // Day of the month without leading zeros 0-31
if($DayOfMonth>21) return 4;
if($DayOfMonth>14) return 3;
if($DayOfMonth>7) return 2;
return 1;
}
From carbon:
return (int) ceil((new Datetime())->format('d') / 7);
As simple as possible :)
Python: Number of the Week in a Month
This is a worked example in Python - should be simple to convert.