check if date is within a date range - php

I have a database filled with a bunch of jobs, where I have startDate and endDate.
What Im trying to do is to list current week and 20 weeks ahead, and see if those weeks are within the database.
And when I echo it out I want the date to be green if the date exists and black if not.
It works, but the problem is that it skips the first week. So if I have 2014-07-01 as startDate and 2014-07-14 as endDate, then the second week in that example will get green, not the first one.
This is what I'v done:
UPDATED
$last = 19;
$today = new \DateTime();
$today->modify('Monday this week');
for ($i=0; $i<$last; $i++) {
$the_week = $today->format('o-m-d');
$sql = mysql_query("SELECT startDate,endDate FROM work WHERE '$the_week' BETWEEN startDate AND endDate'");
$row = mysql_fetch_array($sql);
$startdate = $row['startDate'];
$enddate = $row['endDate'];
if(($startdate > $the_week) AND ($enddate < $the_week)) {
$color = "#69dd54;"; } else { $color = "#ffffff;"; }
echo "<span style='color: ". $color ."'>". $the_week ."</span>";
$new_today->modify('next Monday');
}

What is $the_week? It appears to be a date, but there are many dates in a week. For example there is the first date of a week and the last date of the week. It could be that you have a problematic boundary condition here.
You want the start date to be before the last day of a week, and the end date to be after the first day of a week, I am assuming. Therefore the first week gets excluded because its first day is before the start date, and yet its last day is after the start date.
UPDATE:
Consider having:
$the_week = date('o-m-d', strtotime($today. ' + '. $week .' week'));
$the_week_last = date('o-m-d', strtotime($today. ($week +1) .' week'));
By the way, I would recommend you do a single SQL query for this, rather than one for every week. There is an example of how to do that here.

this is the output of your foreach loop for $the_week
2014-07-09
2014-07-16
2014-07-23
2014-07-30
2014-08-06
2014-08-13
2014-08-20
2014-08-27
2014-09-03
2014-09-10
2014-09-17
2014-09-24
2014-10-01
2014-10-08
2014-10-15
2014-10-22
2014-10-29
2014-11-05
2014-11-12
2014-11-19
as you can see the second one is outside of your date range
Try this as your loop
//$first = 0;
$last = 19;
//$weeks = range($first, $last);
$today = new \DateTime();
$today->modify('Monday this week');
for ($i=0; $i<$last; $i++) {
$the_week = $today->format('o-m-d');
$sql = mysql_query("SELECT startDate,endDate FROM work WHERE $the_week BETWEEN startData AND endDate"); //what happend to your original query.
$r = mysql_num_rows($sql);
$row = mysql_fetch_array($sql);
$startdate = $row['startDate'];
$enddate = $row['endDate'];
if(($startdate > $the_week) AND ($enddate < $the_week)) {
$color = "#69dd54;"; } else { $color = "#ffffff;";
}
echo "<span style='color: ". $color ."'>". $the_week ."</span>";
$today->modify('next Monday');
}
outputs
2014-07-07
2014-07-14
2014-07-21
2014-07-28
2014-08-04
2014-08-11
2014-08-18
2014-08-25
2014-09-01
2014-09-08
2014-09-15
2014-09-22
2014-09-29
2014-10-06
2014-10-13
2014-10-20
2014-10-27
2014-11-03
2014-11-10
Using DateTime you can increment by the 'next Monday' or any day for that matter, with a simple for loop.
also make sure to change
$startdate = $ro['startDate'];
$enddate = $ro['endDate'];
to
$startdate = $row['startDate'];
$enddate = $row['endDate'];

Related

Monthly recurring event on specific date and skip any month if have less no of days compare to user's selected date

I am creating a app where user can create monthly event (Just like Google/Outlook), so if user has selected 31st date of any month and next month have 30 days than all next dates are changing to 30th.
Same goes with 30th, lets say user selected 30th of December than after Feb all next date is changing to 28th.
so if in next month days are lesser than user's selected date its changing the date to that.
Example :
Start Date : 2020-10-31
End Date : 2021-11-30
$diffInMonths = $startDate->diffInMonths($endDate);
for ($i = 0; $i <= $diffInMonths; $i++) {
$newStartDate = $i == 0 ? $startDate : $startDate->addMonthWithNoOverflow();
print('<pre>' . print_r($newStartDate->toDateString(), true) . '</pre>');
}
its giving output like this
2020-10-31
2020-11-30
2020-12-30
2021-01-30
2021-02-28
2021-03-28
2021-04-28
2021-05-28
2021-06-28
2021-07-28
2021-08-28
2021-09-28
2021-10-28
What i need is skip the month if in that month have less days than user;s selected date. With Above example the correct output should be
2020-10-31 //November has less than 31st day
2020-12-31
2021-01-31 //Feb has less than 31st day
2021-03-31 //April has less than 31st day
2021-05-31 //June & July has less than 31st day
2021-07-31
2021-08-31 //September has less than 31st day
2021-10-31
Struggling on this from last 2-3 days so any Kind of help or guidance will made my day :)
This might be dumb way of doing this, but as i am running out of time so here is solution
$dateRange = Carbon::parse($callPlanner->start_date)->toPeriod($callPlanner->end_date, 1, 'month');
$startDate = Carbon::parse($callPlanner->start_date);
foreach ($dateRange as $date) {
// compare the event date with end of month date
if ($startDate->day <= $date->endOfMonth()->day) {
//create new date with this month's year, month and event start date
$newDate = Carbon::createFromDate($date->endOfMonth()->year, $date->endOfMonth()->month, $startDate->day);
echo $newDate->toDateString(). "\n";
}
}
addMonths() add a month to a date. with no overflow, it makes the end of the next month, which is suppose 2020-02-29. in the next step it adds another month to 2020-02-29 and that makes the next date 2020-03-29 because it just added a month. you don't call it to update the date. so your solution would be using lastOfMonth()
$startDate = Carbon::parse('2020-01-31');
$endDate = Carbon::parse('2020-12-31');
$diffInMonths = $startDate->diffInMonths($endDate);
for ($i = 0; $i <= $diffInMonths; $i++) {
$newStartDate = $i == 0 ? $startDate : $startDate->addMonthWithNoOverflow()->lastOfMonth();
print('<pre>' . print_r($newStartDate->toDateString(), true) . '</pre>');
}
which will give you dates like
2020-01-31
2020-02-29
2020-03-31
2020-04-30
2020-05-31
2020-06-30
2020-07-31
2020-08-31
2020-09-30
2020-10-31
2020-11-30
2020-12-31
with addMonth you can't skip months that has less days. you have to check it manually with some condition. like
$startDate = Carbon::parse('2020-10-31');
$endDate = Carbon::parse('2021-11-30');
$highestDate = $startDate->format('d');
$diffInMonths = $startDate->diffInMonths($endDate);
for ($i = 0; $i <= $diffInMonths; $i++) {
$newStartDate = $i == 0 ? $startDate : $startDate->addMonthWithNoOverflow()->lastOfMonth();
if ($newStartDate->format('d') >= $highestDate) {
print('<pre>' . print_r($newStartDate->toDateString(), true) . '</pre>');
}
}
the output is
2020-10-31
2020-12-31
2021-01-31
2021-03-31
2021-05-31
2021-07-31
2021-08-31
2021-10-31
In such cases (similar to monthly subscriptions issues), it could be better to always start from the first date instead of chaining the additions:
$diffInMonths = $startDate->diffInMonths($endDate);
for ($i = 0; $i <= $diffInMonths; $i++) {
$newStartDate = $startDate->copy()->addMonthsWithNoOverflow($i + 1);
print('<pre>' . print_r($newStartDate->toDateString(), true) . '</pre>');
}
This way, you get last day of month is the day is not available but it does not change next iterations.
Note that ->copy() is not needed if you use CarbonImmutable.

Calculate the weeks between 2 dates manually in PHP

I'm trying to calculate the number of months and weeks since a particular date instead of from the beginning of the year.
It shouldn't follow calendar months but should instead count a month as every 4 weeks, and begin from a specified date. I need to be able to display the number of months, and also what week it is (1, 2, 3 or 4).
I want to put in a start date, and have it then count what month and week is it from that start date e.g if the start date is set to Mon 1st August it should show Month 1, Week 1 and so on.
My code is below. I tested it with some different start dates. Here's a list of what the code below generates and what it should display
Jun-20: Should be Week 2 - Shows as Week 0
Jun-27: Should be Week 1 - Shows as Week 3
Jul-04: Should be Week 4 - Shows as Week 2
Jul-11: Should be Week 3 - Shows as Week 1
Jul-18: Should be Week 2 - Shows as Week 0
$monthNumber = 5;
$monthStartDate = '2016-06-13';
$currentStartWeekDate = date('l') != 'Monday' ? date("Y-m-d", strtotime("last monday")) : date("Y-m-d"); // get the current week's Monday's date
$weekDateCounter = $monthStartDate;
$currentWeekNumber = 0;
while ($weekDateCounter != $currentStartWeekDate){
$currentWeekNumber += 1;
$weekDateCounter = date("Y-m-d", strtotime($weekDateCounter . "+7 days"));
//
if ($currentWeekNumber == 4){
$currentWeekNumber = 0; // reset week number
$monthNumber += 1; // increment month number
}
}
I am really at a loss with this and could use any help!
Your approach seems overly complicated:
function weekCounter($startDate,$endDate=null){
//use today as endDate if no date was supplied
$endDate = $endDate? : date('Y-m-d');
//calculate # of full weeks between dates
$secsPerWeek = 60 * 60 * 24 * 7;
$fullWeeks =
floor((strtotime($endDate) - strtotime($startDate))/$secsPerWeek);
$fullMonths = floor($fullWeeks/4);
$weeksRemainder = $fullWeeks % 4; // weeks that don't fit in a month
//increment from 0-base to 1-base, so first week is Week 1. Same with months
$fullMonths++; $weeksRemainder++;
//return months and weeks in an array
return [$fullMonths,$weeksRemainder];
}
You can call the function this way, and capture months and weeks:
//list() will assign the array members from weekCounter to the vars in list
list($months,$weeks) = weekCounter('2016-06-07'); //no end date, so today is used
//now $months and $weeks can be used as you wish
echo "Month: $months, Week: $weeks"; //outputs Month: 2, Week: 2
Live demo
The DateTime classes could make this much simpler for you. Documentation for them is here: http://php.net/manual/en/book.datetime.php
Try this out:
$date1 = new DateTime('2016-04-01');
$date2 = new DateTime('2016-07-24');
$diff = $date1->diff($date2);
$daysInbetween = $diff->days;
$weeksInbetween = floor($diff->days / 7);
$monthsInbetween = floor($weeksInbetween / 4);
print "Days inbetween = $daysInbetween" . PHP_EOL;
print "Weeks inbetween = $weeksInbetween" . PHP_EOL;
print "Months inbetween = $monthsInbetween" . PHP_EOL;
print "Total difference = $monthsInbetween months and "
. ($weeksInbetween - ($monthsInbetween * 4)) . " weeks" . PHP_EOL;
<?php
/**
* AUTHOR : VEDAVITH RAVULA
* DATE : 13122019
*/
function get_weeks($startDate = NULL,$endDate = NULL)
{
if(is_null($startDate) && is_null($endDate))
{
$startDate = date('Y-m-01');
$endDate = date('Y-m-t');
}
$date1 = new DateTime($startDate);
$date2 = new DateTime($endDate);
$interval = $date1->diff($date2);
$weeks = floor(($interval->days) / 7);
if(($date1->format("N") > 1) && ($date1->format("D") != "Sun"))
{
$diffrence = "-".( $date1->format("N"))." Days";
$date1 = $date1->modify($diffrence);
}
for($i = 0; $i <= $weeks; $i++)
{
if($i == 0)
{
$start_date = $date1->format('Y-m-d');
$date1->add(new DateInterval('P6D'));
}
else
{
$date1->add(new DateInterval('P6D'));
}
echo $start_date." - ".$date1->format('Y-m-d')."\n";
$date1->add(new DateInterval('P1D'));
$start_date = $date1->format('Y-m-d');
}
}
//function call
get_weeks("2021-11-01", "2021-11-14");

get current day in next 12 months

Is it possible to get the current day in next 12 months?
Take a look at these scenarios:
Today is 2014-05-09
next month will be 2014-06-09
and so on and so fourth
what if today is 2014-01-31
next month is 2014-02-28
Other examples
what if today is 2014-05-31
next month will be 2014-06-30
thanks
try this.It will work fine even if date is last day of the month also
$date = "2013-03-31";
$d1 = explode("-",$date);
$i = 0;
while ($i < 12)
{
$j = $d1[2];
if(cal_days_in_month(CAL_GREGORIAN, $d1[1], $d1[0]) < $d1[2])
$j = cal_days_in_month(CAL_GREGORIAN, $d1[1], $d1[0]) ;
echo $d1[0]."-".$d1[1]."-".$j."<br>";
$d1[1]++;
if($d1[1] == 13)
{
$d1[0]++;
$d1[1] = 1;
}
$i++;
if($d1[1] < 10)
$d1[1]="0".$d1[1];
}
You can add a DateInterval to the given date. But there is no unified logic in your examples.
For example:
$date = new DateTime( '2014-01-31' );
$aMonthLater = clone $date;
$aMonthLater->add( new DateInterval( 'P1M' ) );
var_dump(
$date->format( DateTime::ISO8601 ),
$aMonthLater->format( DateTime::ISO8601 )
);
This will result in an output like:
2014-01-31T00:00:00+01:00
2014-03-03T00:00:00+01:00
Thus adding a month will actually add a whole month – you example (skipping from 2014-01-31 to 2014-02-28) does not add a full month.
See DateTime::add()

Get week number (in the year) from a date PHP

I want to take a date and work out its week number.
So far, I have the following. It is returning 24 when it should be 42.
<?php
$ddate = "2012-10-18";
$duedt = explode("-",$ddate);
$date = mktime(0, 0, 0, $duedt[2], $duedt[1],$duedt[0]);
$week = (int)date('W', $date);
echo "Weeknummer: ".$week;
?>
Is it wrong and a coincidence that the digits are reversed? Or am I nearly there?
Today, using PHP's DateTime objects is better:
<?php
$ddate = "2012-10-18";
$date = new DateTime($ddate);
$week = $date->format("W");
echo "Weeknummer: $week";
It's because in mktime(), it goes like this:
mktime(hour, minute, second, month, day, year);
Hence, your order is wrong.
<?php
$ddate = "2012-10-18";
$duedt = explode("-", $ddate);
$date = mktime(0, 0, 0, $duedt[1], $duedt[2], $duedt[0]);
$week = (int)date('W', $date);
echo "Weeknummer: " . $week;
?>
$date_string = "2012-10-18";
echo "Weeknummer: " . date("W", strtotime($date_string));
Use PHP's date function
http://php.net/manual/en/function.date.php
date("W", $yourdate)
This get today date then tell the week number for the week
<?php
$date=date("W");
echo $date." Week Number";
?>
Just as a suggestion:
<?php echo date("W", strtotime("2012-10-18")); ?>
Might be a little simpler than all that lot.
Other things you could do:
<?php echo date("Weeknumber: W", strtotime("2012-10-18 01:00:00")); ?>
<?php echo date("Weeknumber: W", strtotime($MY_DATE)); ?>
Becomes more difficult when you need year and week.
Try to find out which week is 01.01.2017.
(It is the 52nd week of 2016, which is from Mon 26.12.2016 - Sun 01.01.2017).
After a longer search I found
strftime('%G-%V',strtotime("2017-01-01"))
Result: 2016-52
https://www.php.net/manual/de/function.strftime.php
ISO-8601:1988 week number of the given year, starting with the first week of the year with at least 4 weekdays, with Monday being the start of the week. (01 through 53)
The equivalent in mysql is DATE_FORMAT(date, '%x-%v')
https://www.w3schools.com/sql/func_mysql_date_format.asp
Week where Monday is the first day of the week (01 to 53).
Could not find a corresponding solution with DateTime.
At least not without solutions like "+1day, last monday".
Edit: since strftime is now deprecated, maybe you can also use date.
Didn't verify it though.
date('o-W',strtotime("2017-01-01"));
I have tried to solve this question for years now, I thought I found a shorter solution but had to come back again to the long story. This function gives back the right ISO week notation:
/**
* calcweek("2018-12-31") => 1901
* This function calculates the production weeknumber according to the start on
* monday and with at least 4 days in the new year. Given that the $date has
* the following format Y-m-d then the outcome is and integer.
*
* #author M.S.B. Bachus
*
* #param date-notation PHP "Y-m-d" showing the data as yyyy-mm-dd
* #return integer
**/
function calcweek($date) {
// 1. Convert input to $year, $month, $day
$dateset = strtotime($date);
$year = date("Y", $dateset);
$month = date("m", $dateset);
$day = date("d", $dateset);
$referenceday = getdate(mktime(0,0,0, $month, $day, $year));
$jan1day = getdate(mktime(0,0,0,1,1,$referenceday[year]));
// 2. check if $year is a leapyear
if ( ($year%4==0 && $year%100!=0) || $year%400==0) {
$leapyear = true;
} else {
$leapyear = false;
}
// 3. check if $year-1 is a leapyear
if ( (($year-1)%4==0 && ($year-1)%100!=0) || ($year-1)%400==0 ) {
$leapyearprev = true;
} else {
$leapyearprev = false;
}
// 4. find the dayofyearnumber for y m d
$mnth = array(0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334);
$dayofyearnumber = $day + $mnth[$month-1];
if ( $leapyear && $month > 2 ) { $dayofyearnumber++; }
// 5. find the jan1weekday for y (monday=1, sunday=7)
$yy = ($year-1)%100;
$c = ($year-1) - $yy;
$g = $yy + intval($yy/4);
$jan1weekday = 1+((((intval($c/100)%4)*5)+$g)%7);
// 6. find the weekday for y m d
$h = $dayofyearnumber + ($jan1weekday-1);
$weekday = 1+(($h-1)%7);
// 7. find if y m d falls in yearnumber y-1, weeknumber 52 or 53
$foundweeknum = false;
if ( $dayofyearnumber <= (8-$jan1weekday) && $jan1weekday > 4 ) {
$yearnumber = $year - 1;
if ( $jan1weekday = 5 || ( $jan1weekday = 6 && $leapyearprev )) {
$weeknumber = 53;
} else {
$weeknumber = 52;
}
$foundweeknum = true;
} else {
$yearnumber = $year;
}
// 8. find if y m d falls in yearnumber y+1, weeknumber 1
if ( $yearnumber == $year && !$foundweeknum) {
if ( $leapyear ) {
$i = 366;
} else {
$i = 365;
}
if ( ($i - $dayofyearnumber) < (4 - $weekday) ) {
$yearnumber = $year + 1;
$weeknumber = 1;
$foundweeknum = true;
}
}
// 9. find if y m d falls in yearnumber y, weeknumber 1 through 53
if ( $yearnumber == $year && !$foundweeknum ) {
$j = $dayofyearnumber + (7 - $weekday) + ($jan1weekday - 1);
$weeknumber = intval( $j/7 );
if ( $jan1weekday > 4 ) { $weeknumber--; }
}
// 10. output iso week number (YYWW)
return ($yearnumber-2000)*100+$weeknumber;
}
I found out that my short solution missed the 2018-12-31 as it gave back 1801 instead of 1901. So I had to put in this long version which is correct.
How about using the IntlGregorianCalendar class?
Requirements: Before you start to use IntlGregorianCalendar make sure that libicu or pecl/intl is installed on the Server.
So run on the CLI:
php -m
If you see intl in the [PHP Modules] list, then you can use IntlGregorianCalendar.
DateTime vs IntlGregorianCalendar:
IntlGregorianCalendar is not better then DateTime. But the good thing about IntlGregorianCalendar is that it will give you the week number as an int.
Example:
$dateTime = new DateTime('21-09-2020 09:00:00');
echo $dateTime->format("W"); // string '39'
$intlCalendar = IntlCalendar::fromDateTime ('21-09-2020 09:00:00');
echo $intlCalendar->get(IntlCalendar::FIELD_WEEK_OF_YEAR); // integer 39
<?php
$ddate = "2012-10-18";
$duedt = explode("-",$ddate);
$date = mktime(0, 0, 0, $duedt[1], $duedt[2],$duedt[0]);
$week = (int)date('W', $date);
echo "Weeknummer: ".$week;
?>
You had the params to mktime wrong - needs to be Month/Day/Year, not Day/Month/Year
To get the week number for a date in North America I do like this:
function week_number($n)
{
$w = date('w', $n);
return 1 + date('z', $n + (6 - $w) * 24 * 3600) / 7;
}
$n = strtotime('2022-12-27');
printf("%s: %d\n", date('D Y-m-d', $n), week_number($n));
and get:
Tue 2022-12-27: 53
for get week number in jalai calendar you can use this:
$weeknumber = date("W"); //number week in year
$dayweek = date("w"); //number day in week
if ($dayweek == "6")
{
$weeknumberint = (int)$weeknumber;
$date2int++;
$weeknumber = (string)$date2int;
}
echo $date2;
result:
15
week number change in saturday
The most of the above given examples create a problem when a year has 53 weeks (like 2020). So every fourth year you will experience a week difference. This code does not:
$thisYear = "2020";
$thisDate = "2020-04-24"; //or any other custom date
$weeknr = date("W", strtotime($thisDate)); //when you want the weeknumber of a specific week, or just enter the weeknumber yourself
$tempDatum = new DateTime();
$tempDatum->setISODate($thisYear, $weeknr);
$tempDatum_start = $tempDatum->format('Y-m-d');
$tempDatum->setISODate($thisYear, $weeknr, 7);
$tempDatum_end = $tempDatum->format('Y-m-d');
echo $tempDatum_start //will output the date of monday
echo $tempDatum_end // will output the date of sunday
Very simple
Just one line:
<?php $date=date("W"); echo "Week " . $date; ?>"
You can also, for example like I needed for a graph, subtract to get the previous week like:
<?php $date=date("W"); echo "Week " . ($date - 1); ?>
Your code will work but you need to flip the 4th and the 5th argument.
I would do it this way
$date_string = "2012-10-18";
$date_int = strtotime($date_string);
$date_date = date($date_int);
$week_number = date('W', $date_date);
echo "Weeknumber: {$week_number}.";
Also, your variable names will be confusing to you after a week of not looking at that code, you should consider reading http://net.tutsplus.com/tutorials/php/why-youre-a-bad-php-programmer/
The rule is that the first week of a year is the week that contains the first Thursday of the year.
I personally use Zend_Date for this kind of calculation and to get the week for today is this simple. They have a lot of other useful functions if you work with dates.
$now = Zend_Date::now();
$week = $now->get(Zend_Date::WEEK);
// 10
To get Correct Week Count for Date 2018-12-31 Please use below Code
$day_count = date('N',strtotime('2018-12-31'));
$week_count = date('W',strtotime('2018-12-31'));
if($week_count=='01' && date('m',strtotime('2018-12-31'))==12){
$yr_count = date('y',strtotime('2018-12-31')) + 1;
}else{
$yr_count = date('y',strtotime('2018-12-31'));
}
function last_monday($date)
{
if (!is_numeric($date))
$date = strtotime($date);
if (date('w', $date) == 1)
return $date;
else
return date('Y-m-d',strtotime('last monday',$date));
}
$date = '2021-01-04'; //Enter custom date
$year = date('Y',strtotime($date));
$date1 = new DateTime($date);
$ldate = last_monday($year."-01-01");
$date2 = new DateTime($ldate);
$diff = $date2->diff($date1)->format("%a");
$diff = $diff/7;
$week = intval($diff) + 1;
echo $week;
//Returns 2.
try this solution
date( 'W', strtotime( "2017-01-01 + 1 day" ) );

Next business day of given date in PHP

Does anyone have a PHP snippet to calculate the next business day for a given date?
How does, for example, YYYY-MM-DD need to be converted to find out the next business day?
Example:
For 03.04.2011 (DD-MM-YYYY) the next business day is 04.04.2011.
For 08.04.2011 the next business day is 11.04.2011.
This is the variable containing the date I need to know the next business day for
$cubeTime['time'];
Variable contains: 2011-04-01
result of the snippet should be: 2011-04-04
Next Weekday
This finds the next weekday from a specific date (not including Saturday or Sunday):
echo date('Y-m-d', strtotime('2011-04-05 +1 Weekday'));
You could also do it with a date variable of course:
$myDate = '2011-04-05';
echo date('Y-m-d', strtotime($myDate . ' +1 Weekday'));
UPDATE: Or, if you have access to PHP's DateTime class (very likely):
$date = new DateTime('2018-01-27');
$date->modify('+7 weekday');
echo $date->format('Y-m-d');
Want to Skip Holidays?:
Although the original poster mentioned "I don't need to consider holidays", if you DO happen to want to ignore holidays, just remember - "Holidays" is just an array of whatever dates you don't want to include and differs by country, region, company, person...etc.
Simply put the above code into a function that excludes/loops past the dates you don't want included. Something like this:
$tmpDate = '2015-06-22';
$holidays = ['2015-07-04', '2015-10-31', '2015-12-25'];
$i = 1;
$nextBusinessDay = date('Y-m-d', strtotime($tmpDate . ' +' . $i . ' Weekday'));
while (in_array($nextBusinessDay, $holidays)) {
$i++;
$nextBusinessDay = date('Y-m-d', strtotime($tmpDate . ' +' . $i . ' Weekday'));
}
I'm sure the above code can be simplified or shortened if you want. I tried to write it in an easy-to-understand way.
For UK holidays you can use
https://www.gov.uk/bank-holidays#england-and-wales
The ICS format data is easy to parse. My suggestion is...
# $date must be in YYYY-MM-DD format
# You can pass in either an array of holidays in YYYYMMDD format
# OR a URL for a .ics file containing holidays
# this defaults to the UK government holiday data for England and Wales
function addBusinessDays($date,$numDays=1,$holidays='') {
if ($holidays==='') $holidays = 'https://www.gov.uk/bank-holidays/england-and-wales.ics';
if (!is_array($holidays)) {
$ch = curl_init($holidays);
curl_setopt($ch,CURLOPT_RETURNTRANSFER,true);
$ics = curl_exec($ch);
curl_close($ch);
$ics = explode("\n",$ics);
$ics = preg_grep('/^DTSTART;/',$ics);
$holidays = preg_replace('/^DTSTART;VALUE=DATE:(\\d{4})(\\d{2})(\\d{2}).*/s','$1-$2-$3',$ics);
}
$addDay = 0;
while ($numDays--) {
while (true) {
$addDay++;
$newDate = date('Y-m-d', strtotime("$date +$addDay Days"));
$newDayOfWeek = date('w', strtotime($newDate));
if ( $newDayOfWeek>0 && $newDayOfWeek<6 && !in_array($newDate,$holidays)) break;
}
}
return $newDate;
}
function next_business_day($date) {
$add_day = 0;
do {
$add_day++;
$new_date = date('Y-m-d', strtotime("$date +$add_day Days"));
$new_day_of_week = date('w', strtotime($new_date));
} while($new_day_of_week == 6 || $new_day_of_week == 0);
return $new_date;
}
This function should ignore weekends (6 = Saturday and 0 = Sunday).
This function will calculate the business day in the future or past. Arguments are number of days, forward (1) or backwards(0), and a date. If no date is supplied todays date will be used:
// returned $date Y/m/d
function work_days_from_date($days, $forward, $date=NULL)
{
if(!$date)
{
$date = date('Y-m-d'); // if no date given, use todays date
}
while ($days != 0)
{
$forward == 1 ? $day = strtotime($date.' +1 day') : $day = strtotime($date.' -1 day');
$date = date('Y-m-d',$day);
if( date('N', strtotime($date)) <= 5) // if it's a weekday
{
$days--;
}
}
return $date;
}
What you need to do is:
Convert the provided date into a timestamp.
Use this along with the or w or N formatters for PHP's date command to tell you what day of the week it is.
If it isn't a "business day", you can then increment the timestamp by a day (86400 seconds) and check again until you hit a business day.
N.B.: For this is really work, you'd also need to exclude any bank or public holidays, etc.
I stumbled apon this thread when I was working on a Danish website where I needed to code a "Next day delivery" PHP script.
Here is what I came up with (This will display the name of the next working day in Danish, and the next working + 1 if current time is more than a given limit)
$day["Mon"] = "Mandag";
$day["Tue"] = "Tirsdag";
$day["Wed"] = "Onsdag";
$day["Thu"] = "Torsdag";
$day["Fri"] = "Fredag";
$day["Sat"] = "Lørdag";
$day["Sun"] = "Søndag";
date_default_timezone_set('Europe/Copenhagen');
$date = date('l');
$checkTime = '1400';
$date2 = date(strtotime($date.' +1 Weekday'));
if( date( 'Hi' ) >= $checkTime) {
$date2 = date(strtotime($date.' +2 Weekday'));
}
if (date('l') == 'Saturday'){
$date2 = date(strtotime($date.' +2 Weekday'));
}
if (date('l') == 'Sunday') {
$date2 = date(strtotime($date.' +2 Weekday'));
}
echo '<p>Næste levering: <span>'.$day[date("D", $date2)].'</span></p>';
As you can see in the sample code $checkTime is where I set the time limit which determines if the next day delivery will be +1 working day or +2 working days.
'1400' = 14:00 hours
I know that the if statements can be made more compressed, but I show my code for people to easily understand the way it works.
I hope someone out there can use this little snippet.
Here is the best way to get business days (Mon-Fri) in PHP.
function days()
{
$week=array();
$weekday=["Monday","Tuesday","Wednesday","Thursday","Friday"];
foreach ($weekday as $key => $value)
{
$sort=$value." this week";
$day=date('D', strtotime($sort));
$date=date('d', strtotime($sort));
$year=date('Y-m-d', strtotime($sort));
$weeks['day']= $day;
$weeks['date']= $date;
$weeks['year']= $year;
$week[]=$weeks;
}
return $week;
}
Hope this will help you guys.
Thanks,.
See the example below:
$startDate = new DateTime( '2013-04-01' ); //intialize start date
$endDate = new DateTime( '2013-04-30' ); //initialize end date
$holiday = array('2013-04-11','2013-04-25'); //this is assumed list of holiday
$interval = new DateInterval('P1D'); // set the interval as 1 day
$daterange = new DatePeriod($startDate, $interval ,$endDate);
foreach($daterange as $date){
if($date->format("N") <6 AND !in_array($date->format("Y-m-d"),$holiday))
$result[] = $date->format("Y-m-d");
}
echo "<pre>";print_r($result);
For more info: http://goo.gl/YOsfPX
You could do something like this.
/**
* #param string $date
* #param DateTimeZone|null|null $DateTimeZone
* #return \NavigableDate\NavigableDateInterface
*/
function getNextBusinessDay(string $date, ? DateTimeZone $DateTimeZone = null):\NavigableDate\NavigableDateInterface
{
$Date = \NavigableDate\NavigableDateFacade::create($date, $DateTimeZone);
$NextDay = $Date->nextDay();
while(true)
{
$nextDayIndexInTheWeek = (int) $NextDay->format('N');
// check if the day is between Monday and Friday. In DateTime class php, Monday is 1 and Friday is 5
if ($nextDayIndexInTheWeek >= 1 && $nextDayIndexInTheWeek <= 5)
{
break;
}
$NextDay = $NextDay->nextDay();
}
return $NextDay;
}
$date = '2017-02-24';
$NextBussinessDay = getNextBusinessDay($date);
var_dump($NextBussinessDay->format('Y-m-d'));
Output:
string(10) "2017-02-27"
\NavigableDate\NavigableDateFacade::create($date, $DateTimeZone), is provided by php library available at https://packagist.org/packages/ishworkh/navigable-date. You need to first include this library in your project with composer or direct download.
I used below methods in PHP, strtotime() does not work specially in leap year February month.
public static function nextWorkingDay($date, $addDays = 1)
{
if (strlen(trim($date)) <= 10) {
$date = trim($date)." 09:00:00";
}
$date = new DateTime($date);
//Add days
$date->add(new DateInterval('P'.$addDays.'D'));
while ($date->format('N') >= 5)
{
$date->add(new DateInterval('P1D'));
}
return $date->format('Y-m-d H:i:s');
}
This solution for 5 working days (you can change if you required for 6 or 4 days working). if you want to exclude more days like holidays then just check another condition in while loop.
//
while ($date->format('N') >= 5 && !in_array($date->format('Y-m-d'), self::holidayArray()))

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