Related
I have a case where a box of text should not be shown between two times for example 20 and 01 (24-hour clock), but it should also work, when i choose not to show the box betweem 20 and 22 for example.
But if I have:
$start = "20";
$end = "01";
$now = date('H');
if($now > $start AND $now < $end) {
echo "DONT SHOW THE BOX";
} else {
echo "SHOW THE BOX";
}
How can I convert the numbers, can I use mktime() even if I don't have a date? Because the box should be activated every day in that time range.
You don't want to show the box between 20:00 to 01:00
Currently your logic is kinda messed up in if($now > $start AND $now < $end).
If you expect $start = 20 and $end = 1, Then what kind of value is $now that might be MORE than 20 AND LESS than 1.
Your if statement logic will always go to ELSE whatever the value of $now is.
But there's another workaround to switch the logic like this.
You want to show the box between 02:00 to 19:00
Instead of the other way around.
So you can do this,
$start = "20";
$end = "01";
$now = date('H');
if ($now > $end && $now < $start)
{
echo "SHOW THE BOX";
}
else
{
echo "DON'T SHOW";
}
Update 1:
Now, you don't want to show the box between 20:00 to 22:00
You can do the vice versa or which is your current logic. Like,
$start = "20";
$end = "22";
$now = date('H');
if ($now >= $start && $now <= $end)
{
echo "DON'T SHOW";
}
else
{
echo "SHOW THE BOX";
}
Update 2:
If the $start or $end varies, you can always wrap them in another if condition. Like,
if ($start > $end)
{
if ($now > $end && $now < $start)
{
echo "SHOW THE BOX";
}
else
{
echo "DON'T SHOW";
}
}
else if ($start < $end)
{
if ($now >= $start && $now <= $end)
{
echo "DON'T SHOW";
}
else
{
echo "SHOW THE BOX";
}
}
The code isn't very pretty, but I think this is what you are after?
define('START_TIME', 6);
define('END_TIME', 1);
$startTime = new DateTime();
$startTime->setTime(START_TIME, 0);
$endTime = clone $startTime;
if (START_TIME > END_TIME) {
$endTime->modify('+1 day');
}
$endTime->setTime(END_TIME, 0);
$currentTime = new DateTime();
if ($currentTime > $startTime && $currentTime < $endTime) {
echo 'Show box';
} else {
echo 'Don\'t show box';
}
You can alter the hours to decide the start/end hour for when to show/hide. If the ending hour is a lower hour than the starting hour then it will presume the ending hour should be the next day.
It could potentially be cleaned up a bit, but I'll leave that in your hands if it is what you're after.
date_default_timezone_set("Europe/Stockholm");
$setdate = ['14-05-2022','15-05-2022','16-05-2022'];
print_r($setdate)."<br>";
foreach ($setdate as $value) {
$date= date("d-m-Y");
if($date == $value){
// echo "set Data For Product";
// die();
}
}
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 to check if the current daytime falls in a specific range. I looked up the internet and found several similar solutions like this one:
$now = date("His");//or date("H:i:s")
$start = '130000';//or '13:00:00'
$end = '170000';//or '17:00:00'
if($now >= $start && $now <= $end){
echo "Time in between";
}
else{
echo "Time outside constraints";
}
If both conditions have to be true, how can this bis achieved when we assume that $start is 06:00:00 and $end is 02:00:00.
If we make the assumption that it is 01:00:00, in this case the first condition can't be true.
Has anybody an idea to handle this problem differently?
Thanks!
Naturally, you'd have to account for date in your comparisons.
<?php
$start = strtotime('2014-11-17 06:00:00');
$end = strtotime('2014-11-18 02:00:00');
if(time() >= $start && time() <= $end) {
// ok
} else {
// not ok
}
If you need to check whether or not the time frame rolls over midnight
function isWithinTimeRange($start, $end){
$now = date("His");
// time frame rolls over midnight
if($start > $end) {
// if current time is past start time or before end time
if($now >= $start || $now < $end){
return true;
}
}
// else time frame is within same day check if we are between start and end
else if ($now >= $start && $now <= $end) {
return true;
}
return false;
}
You can then get whether or not you are within that time frame by
echo isWithinTimeRange(130000, 170000);
date_default_timezone_set("Asia/Colombo");
$nowDate = date("Y-m-d h:i:sa");
//echo '<br>' . $nowDate;
$start = '21:39:35';
$end = '25:39:35';
$time = date("H:i:s", strtotime($nowDate));
$this->isWithInTime($start, $end, $time);
function isWithInTime($start,$end,$time) {
if (($time >= $start )&& ($time <= $end)) {
// echo 'OK';
return TRUE;
} else {
//echo 'Not OK';
return FALSE;
}
}
Cannot comment due to low reputation, but #DOfficial answer is great but be aware of inconsistency in comparision.
Original
// if current time is past start time or before end time
if($now >= $start || $now < $end){
Should be imho
// if current time is past start time or before end time
if($now >= $start || $now <= $end){
I have array of week numbers from 1 to 52. how i can convert it to
[week 1 jan],[week 2 jan] .......
using PHP
OK ... I fix it and here is my code
function getWeeks($date, $rollover)
{
$cut = substr($date, 0, 8);
$daylen = 86400;
$timestamp = strtotime($date);
$first = strtotime($cut . "00");
$elapsed = ($timestamp - $first) / $daylen;
$i = 1;
$weeks = 1;
for($i; $i<=$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;
}
and in the foreach I added
$x="1/1/2013 + ".$record->tms." weeks";
$m=date("Y-m-d", strtotime($x));
$first_week_start=getWeeks($m, "sunday");
if($first_week_start == 1){$typo="st";}
if($first_week_start == 2){$typo="nd";}
if($first_week_start == 3){$typo="rd";}
if($first_week_start == 4){$typo="th";}
if($first_week_start == 5){$typo="th";}
$month=date("M", strtotime($m));
$final_format .= "'".$first_week_start.$typo ." week in ".$month."'";
Try a loop, start with 1/jan, use DateInterval::createFromDateString('1 week'); and DateTime::add each time in the loop to add the next week, use DateTime::format to get the month also checking the current year in each iteration to make sure the loop hasn't moved to the next year.
Code:
date_default_timezone_set("UTC");
$weeks = array();
$dt = new DateTime("2013-01-01");
$interval = DateInterval::createFromDateString("1 week");
for($i=1; $i <= 52; $i++)
{
$weeks[$i] = "week " . $i . " " . $dt->format("M");
$dt = $dt->add($interval);
}
print_r($weeks);
If you want a function that will return a month abbreviation from a week:
function weekMonth($week)
{
date_default_timezone_set("UTC");
return (new DateTime("2013-01-01"))->add(DateInterval::createFromDateString($week." week"))->format("M");
}
echo weekMonth(6);
I want to calculate the number of weekdays days in a give month and year. Weekdays means monday to friday. How do i do it ?
You don't need to count every day in the month. You already know the first 28 days contain 20 weekdays no matter what. All you have to do is determine the last few days. Change the start value to 29. Then add 20 weekdays to your return value.
function get_weekdays($m,$y) {
$lastday = date("t",mktime(0,0,0,$m,1,$y));
$weekdays=0;
for($d=29;$d<=$lastday;$d++) {
$wd = date("w",mktime(0,0,0,$m,$d,$y));
if($wd > 0 && $wd < 6) $weekdays++;
}
return $weekdays+20;
}
Some basic code:
$month = 12;
$weekdays = array();
$d = 1;
do {
$mk = mktime(0, 0, 0, $month, $d, date("Y"));
#$weekdays[date("w", $mk)]++;
$d++;
} while (date("m", $mk) == $month);
print_r($weekdays);
Remove the # if your PHP error warning doesn't show notices.
try this one
function getWeekdays($m, $y = NULL){
$arrDtext = array('Mon', 'Tue', 'Wed', 'Thu', 'Fri');
if(is_null($y) || (!is_null($y) && $y == ''))
$y = date('Y');
$d = 1;
$timestamp = mktime(0,0,0,$m,$d,$y);
$lastDate = date('t', $timestamp);
$workingDays = 0;
for($i=$d; $i<=$lastDate; $i++){
if(in_array(date('D', mktime(0,0,0,$m,$i,$y)), $arrDtext)){
$workingDays++;
}
}
return $workingDays;
}
This is the simplest code I could come up with.
You really would need to create an array or a database table to hold the holidays to get a true, "Working Days" count, but that wasn't what was asked, so here you go, hope this helps someone.
function get_weekdays($m,$y) {
$lastday = date("t",mktime(0,0,0,$m,1,$y));
$weekdays=0;
for($d=1;$d<=$lastday;$d++) {
$wd = date("w",mktime(0,0,0,$m,$d,$y));
if($wd > 0 && $wd < 6) $weekdays++;
}
return $weekdays;
}
Get the number of working days without holidays between two dates :
Use example:
echo number_of_working_days('2013-12-23', '2013-12-29');
Output:
3
Link to the function
DateObject method:
function getWorkingDays(DateTime $date) {
$month = clone $date;
$month->modify('last day of this month');
$workingDays = 0;
for ($i = $month->format('t'); $i > 28; --$i) {
if ($month->format('N') < 6) {
++$workingDays;
}
$month->modify('-1 day');
}
return 20 + $workingDays;
}
Calculate working days in a month from any date:
public function getworkd($mday)
{
$dn = new DateTime($mday);
$dfrom = $dn->format('Y-m-01');
$dtill = $dn->format('Y-m-t');
$df = new DateTime($dfrom);
$dt = new DateTime($dtill);
$wdays = 0;
while($df<=$dt)
{
$dof= $df->format('D') ;
if( $dof == 'Sun' || $dof == 'Sat' ) ; else $wdays++;
$df->add(new DateInterval('P1D'));
}
return $wdays;
}
Find the last day and the weekday for the given month
then do a simple while loop like :-
$dates = explode(',', date('t,N', strtotime('2013-11-01')));
$day = $dates[1];
$tot = $dates[0];
$cnt = 0;
while ($tot>1)
{
if ($day < 6)
{
$cnt++;
}
if ($day == 1)
{
$day = 7;
}
else
{
$day--;
}
$tot--;
}
$cnt = total of weekday (Monday to Friday) for a given month
I've come up with a non-loop function. Much better in terms of performance. It might seem messy but it just needs to ask PHP the first day's weekday and the month's number days: the rest are arithmetical operations based on logic.
function countWorkDays($year, $month)
{
$workingWeekdays = 5;
$firstDayTimestamp = mktime(0, 0, 0, $month, 1, $year);
$firstDayWeekDay = (int)date("N", $firstDayTimestamp); //1: monday, 7: saturday
$upToDay = (int)date("t", $firstDayTimestamp);
$firstMonday = 1 === $firstDayWeekDay ? 1 : 9 - $firstDayWeekDay;
$wholeWeeks = $firstMonday < $upToDay ? (int)floor(($upToDay - $firstMonday + 1) / 7) : 0;
$extraDays = ($upToDay - $firstMonday + 1) % 7;
$initialWorkdays = $firstMonday > 1 && $firstDayWeekDay <= $workingWeekdays ? $workingWeekdays - $firstDayWeekDay + 1 : 0;
$workdaysInWholeWeeks = $wholeWeeks * $workingWeekdays;
$extraWorkdays = $extraDays <= $workingWeekdays ? $extraDays : $workingWeekdays;
return $initialWorkdays + $workdaysInWholeWeeks + $extraWorkdays;
}
These functions work Without Loops.
The functions calculate the number of weekdays using:
day-number of first monday in month
number of days in month
// main functions
// weekdays in month of year
function calculateNumberOfWeekDaysAtDate($month, $year)
{
// I'm sorry, I don't know the right format for the $month and $year, I hope this is right.
// PLEASE CORRECT IF WRONG
$firstMondayInCurrentMonth = (int) date("j", strtotime("first monday of 01-$month-$year")); //get first monday in month for calculations
$numberOfDaysOfCurrentMonth = (int) date("t", strtotime("01-$month-$year")); // number of days in month
return calculateNumberOfWeekDaysFromFirstMondayAndNumberOfMonthDays($firstMondayInCurrentMonth, $numberOfDaysOfCurrentMonth);
}
// week days in current month
function calculateNumberOfWeekDaysInCurrentMonth()
{
$firstMondayInCurrentMonth = (int) date("j", strtotime("first monday of this month")); //get first monday in month for calculations
$numberOfDaysOfCurrentMonth = (int) date("t"); // number of days in this month
return calculateNumberOfWeekDaysFromFirstMondayAndNumberOfMonthDays($firstMondayInCurrentMonth, $numberOfDaysOfCurrentMonth);
}
// helper functions
function calculateNumberOfWeekDaysFromFirstMondayAndNumberOfMonthDays($firstMondayInCurrentMonth, $numberOfDaysOfCurrentMonth)
{
return $numberOfWeekDays = (($start = ($firstMondayInCurrentMonth - 3)) < 0 ? 0 : $start) + floor(($numberOfDaysOfCurrentMonth - ($firstMondayInCurrentMonth - 1)) / 7) * 5 + (($rest = (($numberOfDaysOfCurrentMonth - ($firstMondayInCurrentMonth - 1)) % 7)) <= 5 ? $rest : 5);
}
function workingDays($m,$y) {
$days = cal_days_in_month(CAL_GREGORIAN, $m, $y);
$workig_days = 0;
$days_rest = array(5,6); //friday,saturday
for ( $d=1 ; $d < $days+1 ; $d++ ) {
if ( !in_array(date("w",strtotime("{$d}-{$m}-{$y}")),$days_rest) ) {
$workig_days++;
}
}
return $workig_days;
}
I created a simple function that takes the $first_day_of_month (week day like Sunday/Monday etc). You can find out the first day of month like this:
date('N', strtotime(date("01-m-Y")));
And using the $month_last_date which can be procured like this:
date("t");
Here is the function:
function workingDaysInMonth(int $first_day_of_month, int $month_last_date) : array
{
$working_days = [];
$day = $first_day_of_month;
$working_day_count = 0;
for ($i = 1; $i <= $month_last_date; $i++) {
if ($day == 8) {
$day = 1;
}
if (!($day == 6 || $day == 7)) {
$working_day_count++;
$working_days[$i] = $working_day_count;
}
$day++;
}
return $working_days;
}
this will work
// oct. 2013
$month = 10;
// loop through month days
for ($i = 1; $i <= 31; $i++) {
// given month timestamp
$timestamp = mktime(0, 0, 0, $month, $i, 2012);
// to be sure we have not gone to the next month
if (date("n", $timestamp) == $month) {
// current day in the loop
$day = date("N", $timestamp);
// if this is between 1 to 5, weekdays, 1 = Monday, 5 = Friday
if ($day == 1 OR $day <= 5) {
// write it down now
$days[$day][] = date("j", $timestamp);
}
}
}
// to see if it works :)
print_r($days);