Okay first of all, before some of you might put this question into a duplicate question. i have been go through many questions related to my question. But it seems like those questions cannot solve my problem. i want to find days between 2 dates excluding weekend and holidays. the questions that i have been go through declare the variable of $holiday in array and initialize the value in that array.
Below is my code in LeaveapplicationController.phpfor finding days exclude weekends
$date1 = $model->startDate;
$date2 = $model->endDate;
$date1 = strtotime($date1);
$date2 = strtotime($date2);
//Initialized public holiday
$holidays = array("2020-01-21", "2020-01-22", "2020-01-23");
$days = ($date2 - $date1)/86400 + 1;
$fullWeek = floor($days/7);
$remainDay = fmod($days,7);
$firstDay = date("N", $date1);
$lastDay = date("N", $date2);
if($firstDay <= $lastDay){
if($firstDay <= 6 && 6 <= $lastDay) $remainDay--;
if($firstDay <= 7 && 7 <= $lastDay) $remainDay--;
}
else
{
if($firstDay == 7){
$remainDay--;
if($lastDay == 6){
$remainDay--;
}
}
else{
$remainDay -= 2;
}
}
$workDay = $fullWeek * 5;
if($remainDay > 0)
{
$workDay += $remainDay;
}
foreach($holidays as $holiday){
$timeStamp = strtotime($holiday);
if($date1 <= $timeStamp && $timeStamp <= $date2 && date("N", $timeStamp) != 6 && date("N", $timeStamp) != 7)
$workDay--;
}
$model->no_of_days=$workDay;
$model->save();
So, i have tried with holiday that has been initialized with some values. In my case, i want to replace that $holiday=array() with data from another table of database. Is there any best way to do it?
Using DateTime
$start = new \DateTime($model->startDate); //2020-01-01
$end = new \DateTime($model->endDate); //2020-01-31
$endDate = $end->format('Y-m-d');
$interval = new \DateInterval('P1D');
$end->add($interval);
$period = new \DatePeriod($start, $interval, $end);
foreach ($period as $date) {
$allDates[] = [
'date' => $date->format('Y-m-d'),
'dayNo' => $date->format('N'),
];
}
//Initialized public holiday
$holidays = HolidayModelName::find()
->select('date')
->where(['between', 'date', $start->format('Y-m-d'), $endDate])
->indexBy('date')
->column();
$workDay = 0;
foreach ($allDates as $value) {
$isWeekOff = $value['dayNo'] == 6 || $value['dayNo'] == 7;
if (!$isWeekOff && !isset($holidays[$value['date']])) {
$workDay++;
}
}
// Result : 20 Work Days
Anyone know, how I can achieve below:
If someone select the option "For weekend only" then all the weekend dates will be selected from the current month to the next month's specified date.
if someone select the option "For weekday only" then all the weekday dates will be selected from the current month to the next month's specified date.
if someone select the option "For Friday only" then all the Friday dates will be selected from the current month to the next month's specified date.
Something like above:
Anyone have any idea where I can start this with??
$('#dt_1').datepicker({
rtl: KTUtil.isRTL(),
todayHighlight: true,
templates: arrows,
startDate: date, //disable all old dates
setDate: date, //tomorrow's date allowed
multidate: true,
format: 'dd/mm/yyyy'
});
I've found the solution to achieve this with the help of another post of stackoverflow:
get weekend using php
PHP code:
//weekend.....
$now = strtotime("now");
$end_date = strtotime("+3 weeks");
$weekend_array = array();
while (date("d-m-Y", $now) != date("d-m-Y", $end_date)) {
$day_index = date("w", $now);
if ($day_index == 0 || $day_index == 6) {
// Print or store the weekends here.
//echo "<br>".$day_index."- Date: ".date("d-m-Y", $now);
$weekend_array[] = date("d-m-Y", $now);
}
$now = strtotime(date("d-m-Y", $now) . "+1 day");
}
//friday only....
$now = strtotime("now");
$end_date = strtotime("+3 weeks");
$friday_array = array();
while (date("d-m-Y", $now) != date("d-m-Y", $end_date)) {
$day_index = date("w", $now);
if ($day_index == 5) {
// Print or store the weekends here.
//echo "<br>".$day_index."- Date: ".date("d-m-Y", $now);
$friday_array[] = date("d-m-Y", $now);
}
$now = strtotime(date("d-m-Y", $now) . "+1 day");
}
//weekdays......
$now = strtotime("now");
$end_date = strtotime("+3 weeks");
$weekday_array = array();
while (date("d-m-Y", $now) != date("d-m-Y", $end_date)) {
$day_index = date("w", $now);
if ($day_index >= 1 && $day_index <= 5) {
// Print or store the weekends here.
//echo "<br>".$day_index."- Date: ".date("d-m-Y", $now);
$weekday_array[] = date("d-m-Y", $now);
}
$now = strtotime(date("d-m-Y", $now) . "+1 day");
}
Javascript code:
$(document).ready( function () {
$("#sel_dropdown").change(function() {
$( "select option:selected" ).each(function() {
var v = $(this).val();
if(v == 1) {
//weekend only....
$('#dt_1').datepicker('setDates',[<?php
$i=0;
for($i=0;$i<=count($weekend_array);$i++) {
echo "'".$weekend_array[$i]."',";
}
?>]);
}
if(v == 2) {
//Friday only.....
$('#dt_1').datepicker('setDates',[<?php
$i=0;
for($i=0;$i<=count($friday_array);$i++) {
echo "'".$friday_array[$i]."',";
}
?>]);
}
if(v == 3) {
//Weekdays only....
$('#dt_1').datepicker('setDates',[<?php
$i=0;
for($i=0;$i<=count($weekday_array);$i++) {
echo "'".$weekday_array[$i]."',";
}
?>]);
}
if(v == 4) {
//custom option....
$('#dt_1').data('datepicker').setDate(null);
}
});
});
});
</script>
Hopefully, this will useful for anyone looking for this type of solution.
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;
$weekNumber = getWeekNumber('2015-09-17') //output 3;
I think this relationship should be true and come in handy:
Week of the month = Week of the year - Week of the year of first day of month + 1
We also need to make sure that "overlapping" weeks from the previous year are handeled correctly - if January 1st is in week 52 or 53, it should be counted as week 0. In a similar fashion, if a day in December is in the first week of the next year, it should be counted as 53. (Previous versions of this answer failed to do this properly.)
<?php
function weekOfMonth($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;
}
}
// 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", ...).
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 $w;
}
The corect way is
function weekOfMonth($date) {
$firstOfMonth = date("Y-m-01", strtotime($date));
return intval(date("W", strtotime($date))) - intval(date("W", strtotime($firstOfMonth)));
}
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;
function getWeekOfMonth(DateTime $date) {
$firstDayOfMonth = new DateTime($date->format('Y-m-1'));
return ceil(($firstDayOfMonth->format('N') + $date->format('j') - 1) / 7);
}
Goendg solution does not work for 2016-10-31.
function weekOfMonth($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
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;
}
You can also use this simple formula for finding week of the month
$currentWeek = ceil((date("d",strtotime($today_date)) - date("w",strtotime($today_date)) - 1) / 7) + 1;
ALGORITHM :
Date = '2018-08-08' => Y-m-d
Find out day of the month eg. 08
Find out Numeric representation of the day of the week minus 1 (number of days in week) eg. (3-1)
Take difference and store in result
Subtract 1 from result
Divide it by 7 to result and ceil the value of result
Add 1 to result eg. ceil(( 08 - 3 ) - 1 ) / 7) + 1 = 2
My function. The main idea: we would count amount of weeks passed from the month's first date to current. And the current week number would be the next one. Works on rule: "Week starts from monday" (for sunday-based type we need to transform the increasing algorithm)
function GetWeekNumberOfMonth ($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);
$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 '<pre>'; print_r($weeks_ar);
working fine.
// 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';
// 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;
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!
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;
$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: ". $week_no;
How about this function making use of PHP's relative dates?
This function assumes the week ends on Saturday. But this can be changed easily.
function get_weekNumMonth($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;
}
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
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;
}
//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
I have an array structured like this:
Array ( [0] => 24-12-2013 [1] => 25-12-2013 [2] => 26-12-2014 [3] => 27-12-2013 [4])
I would like to check if any of the dates in the array are within a given date range.
The date range is structured like this:
$start = (date("d-m-Y", strtotime('25-12-2013')));
$end = (date("d-m-Y", strtotime('26'12'2013')));
I would like to know which dates in the array are within the date range.
Couple things:
Use timestamps or DateTime objects to compare dates, not strings
Use date format YYYY-MM-DD to avoid potential ambiguity about your date format (d/m/y or m/d/y)
This code will do what you want:
$dates = array("2013-12-24","2013-12-25","2014-12-24","2013-12-27");
$start = strtotime('2013-12-25');
$end = strtotime('2013-12-26');
foreach($dates AS $date) {
$timestamp = strtotime($date);
if($timestamp >= $start && $timestamp <= $end) {
echo "The date $date is within our date range\n";
} else {
echo "The date $date is NOT within our date range\n";
}
}
See it in action:
http://3v4l.org/GWJI2
$dates = array ('24-12-2013', '25-12-2013', '26-12-2014', '27-12-2013');
$start = strtotime('25-12-2013');
$end = strtotime('26-12-2013');
$inDateRange = count(
array_filter(
$dates,
function($value) use($start, $end) {
$value = strtotime($value);
return ($value >= $start && $value <= $end);
}
)
);
<?php
$start = DateTime::createFromFormat('d-m-Y', '25-12-2013');
$end = DateTime::createFromFormat('d-m-Y', '26-12-2013');
$dates = array('24-12-2013','25-12-2013','26-12-2014','27-12-2013');
$matches = array();
foreach ($dates as $date) {
$date2 = DateTime::createFromFormat('d-m-Y', $date);
if ($date2 >= $start && $date2 =< $end) {
$matches[] = $date;
}
}
print_r($matches);
See it in action
$_between = array();
$start = date('Ymd', strtotime($start));
$end = date('Ymd', strtotime($end));
foreach ($dates as $date)
{
$date = date('Ymd',strtotime($date));
if ($date > $start && $date < $end) {
array_push($_between,$date);
continue;
}
}
echo '<pre>';
var_dump($_between);
echo '</pre>';
Loop over the array turning each date into unix time (seconds since Jan 1, 1970), and do simple math to see if the number of seconds is between the range. Like so:
$start = strtotime('25-12-2013');
$end = strtotime('26'12'2013');
foreach($date in $dates) {
$unix_time = strtotime($date);
if($unix_time > $start && $unix_time < $end)
//in range
}
// PHP >= 5.3:
$dates_in_range = array_filter($array, function($date) {
global $start;
global $end;
return (strtotime($date) >= strtotime($start) and strtotime($date) <= strtotime($end));
});