I have this function witch return an array of date. I need to jump on every seven days from now until last year.
$date[] = $lastDate = (new \DateTIme('NOW'))->format('Y-m-d');
for ($i = 1; $i < 54; ++$i) { // 54 -> number of weeks in a year
$date[] = $lastDate = date('Y-m-d', strtotime('-7 day', strtotime($lastDate)));
}
return array_reverse($date);
It works but I can do better.
I would like to change it because using 54 for the number of weeks in a year is not very good. (it can change)
So I want to use the DateInterval php class.
I can have the date of the last year with :
$lastYear = date('Y-m-d', strtotime('-1 year', strtotime($lastDate)));
But I don't know how I can have my array with all my dates with the DateInterval class.
Can someone help me? I'm very bad with date manipulation :( ...
Here is an example array about what I need:
["2015-07-06", "2015-07-13", "2015-07-20", "2015-07-27", "2015-08-03", "2015-08-10", "2015-08-17", "2015-08-24", "2015-08-31", "2015-09-07", "2015-09-14", "2015-09-21", "2015-09-28", "2015-10-05", "2015-10-12", "2015-10-19", "2015-10-26", "2015-11-02", "2015-11-09", "2015-11-16", "2015-11-23", "2015-11-30", "2015-12-07", "2015-12-14", "2015-12-21", "2015-12-28", "2016-01-04", "2016-01-11", "2016-01-18", "2016-01-25", "2016-02-01", "2016-02-08", "2016-02-15", "2016-02-22", "2016-02-29", "2016-03-07", "2016-03-14", "2016-03-21", "2016-03-28", "2016-04-04", "2016-04-11", "2016-04-18", "2016-04-25", "2016-05-02", "2016-05-09", "2016-05-16", "2016-05-23", "2016-05-30", "2016-06-06", "2016-06-13", "2016-06-20", "2016-06-27", "2016-07-04"]
PHP got it 's own native DateInterval object. Here 's a short example how to use it.
$oPeriodStart = new DateTime();
$oPeriodEnd = new DateTime('+12 months');
$oPeriod = new DatePeriod(
$oPeriodStart,
DateInterval::createFromDateString('7 days'),
$oPeriodEnd
);
foreach ($oPeriod as $oInterval) {
var_dump($oInterval->format('Y-m-d));
}
So what we 've done here? For a period of dates you need a start date, an end date and the interval. Just test it for yourself. Have fun.
Try this:
$timestamp = strtotime("last Sunday");
$sundays = array();
$last_year_timestamp = strtotime("-1 year ",$timestamp);
while($timestamp >= $last_year_timestamp) {
if (date("w", $timestamp) == 0) {
$sundays[] = date("Y-m-d", $timestamp);
$timestamp -= 86400*7;
continue;
}
$timestamp -= 86400;
}
Related
How can I compare two dates in PHP?
The date is stored in the database in the following format
2011-10-2
If I wanted to compare today's date against the date in the database to see which one is greater, how would I do it?
I tried this,
$today = date("Y-m-d");
$expire = $row->expireDate //from db
if($today < $expireDate) { //do something; }
but it doesn't really work that way. What's another way of doing it?
If all your dates are posterior to the 1st of January of 1970, you could use something like:
$today = date("Y-m-d");
$expire = $row->expireDate; //from database
$today_time = strtotime($today);
$expire_time = strtotime($expire);
if ($expire_time < $today_time) { /* do Something */ }
If you are using PHP 5 >= 5.2.0, you could use the DateTime class:
$today_dt = new DateTime($today);
$expire_dt = new DateTime($expire);
if ($expire_dt < $today_dt) { /* Do something */ }
Or something along these lines.
in the database the date looks like this 2011-10-2
Store it in YYYY-MM-DD and then string comparison will work because '1' > '0', etc.
Just to compliment the already given answers, see the following example:
$today = new DateTime('');
$expireDate = new DateTime($row->expireDate); //from database
if($today->format("Y-m-d") < $expireDate->format("Y-m-d")) {
//do something;
}
Update:
Or simple use old-school date() function:
if(date('Y-m-d') < date('Y-m-d', strtotime($expire_date))){
//echo not yet expired!
}
I would'nt do this with PHP.
A database should know, what day is today.( use MySQL->NOW() for example ), so it will be very easy to compare within the Query and return the result, without any problems depending on the used Date-Types
SELECT IF(expireDate < NOW(),TRUE,FALSE) as isExpired FROM tableName
$today = date('Y-m-d');//Y-m-d H:i:s
$expireDate = new DateTime($row->expireDate);// From db
$date1=date_create($today);
$date2=date_create($expireDate->format('Y-m-d'));
$diff=date_diff($date1,$date2);
//echo $timeDiff;
if($diff->days >= 30){
echo "Expired.";
}else{
echo "Not expired.";
}
Here's a way on how to get the difference between two dates in minutes.
// set dates
$date_compare1= date("d-m-Y h:i:s a", strtotime($date1));
// date now
$date_compare2= date("d-m-Y h:i:s a", strtotime($date2));
// calculate the difference
$difference = strtotime($date_compare1) - strtotime($date_compare2);
$difference_in_minutes = $difference / 60;
echo $difference_in_minutes;
You can convert the dates into UNIX timestamps and compare the difference between them in seconds.
$today_date=date("Y-m-d");
$entered_date=$_POST['date'];
$dateTimestamp1 = strtotime($today_date);
$dateTimestamp2 = strtotime($entered_date);
$diff= $dateTimestamp1-$dateTimestamp2;
//echo $diff;
if ($diff<=0)
{
echo "Enter a valid date";
}
I had that problem too and I solve it by:
$today = date("Ymd");
$expire = str_replace('-', '', $row->expireDate); //from db
if(($today - $expire) > $NUMBER_OF_DAYS)
{
//do something;
}
Here's my spin on how to get the difference in days between two dates with PHP.
Note the use of '!' in the format to discard the time part of the dates, thanks to info from DateTime createFromFormat without time.
$today = DateTime::createFromFormat('!Y-m-d', date('Y-m-d'));
$wanted = DateTime::createFromFormat('!d-m-Y', $row["WANTED_DELIVERY_DATE"]);
$diff = $today->diff($wanted);
$days = $diff->days;
if (($diff->invert) != 0) $days = -1 * $days;
$overdue = (($days < 0) ? true : false);
print "<!-- (".(($days > 0) ? '+' : '').($days).") -->\n";
Found the answer on a blog and it's as simple as:
strtotime(date("Y"."-01-01")) -strtotime($newdate))/86400
And you'll get the days between the 2 dates.
This works because of PHP's string comparison logic. Simply you can check...
if ($startdate < $date) {// do something}
if ($startdate > $date) {// do something}
Both dates must be in the same format. Digits need to be zero-padded to the left and ordered from most significant to least significant. Y-m-d and Y-m-d H:i:s satisfy these conditions.
If you want a date ($date) to get expired in some interval for example a token expiration date when performing a password reset, here's how you can do:
$date = $row->expireDate;
$date->add(new DateInterval('PT24H')); // adds 24 hours
$now = new \DateTime();
if($now < $date) { /* expired after 24 hours */ }
But in your case you could do the comparison just as the following:
$today = new DateTime('Y-m-d');
$date = $row->expireDate;
if($today < $date) { /* do something */ }
first of all, try to give the format you want to the current date time of your server:
Obtain current date time
$current_date = getdate();
Separate date and time to manage them as you wish:
$current_date_only = $current_date[year].'-'.$current_date[mon].'-'.$current_date[mday];
$current_time_only = $current_date['hours'].':'.$current_date['minutes'].':'.$current_date['seconds'];
Compare it depending if you are using donly date or datetime in your DB:
$today = $current_date_only.' '.$current_time_only;
or
$today = $current_date_only;
if($today < $expireDate)
hope it helps
I want to add 12 months to my date. My start date is 02/29/2020 and I want to add 12 months to this.
Code:
$startdate = '02/29/2020';
date('m/d/Y', strtotime('+12 months', strtotime($startdate)));
Output:
03/01/2021
I used this code to add 12 months but the output is 03/01/2021, when the real output should be 02/28/2020.
Have a look!
function add_months($months, DateTime $dateObject)
{
$next = new DateTime($dateObject->format('Y-m-d'));
$next->modify('last day of +'.$months.' month');
if($dateObject->format('d') > $next->format('d')) {
return $dateObject->diff($next);
} else {
return new DateInterval('P'.$months.'M');
}
}
function getCalculatedDate($d1, $months)
{
$date = new DateTime($d1);
// call second function to add the months
$newDate = $date->add(add_months($months, $date));
//formats final date to m/d/Y form
$dateReturned = $newDate->format('m/d/Y');
return $dateReturned;
}
An example would be:-
$startDate = '02/29/2020';
$nMonths = 12; // choose how many months you want to add
$finalDate = getCalculatedDate($startDate, $nMonths); // output: 02/28/2021
This way you will get the output of 02/28/2021
$startdate = '02/29/2020';
$date = date('m/d/Y', strtotime($startdate . '+365 days'));
using DateTime and DateInterval objects leads to 03/01/2021
$date = new \DateTime('02/29/2020');
$date->add(new \DateInterval('P12M'));
echo $date->format('m/d/Y');
for me 03/01/2021 is not always a bad answer
How can I compare two dates in PHP?
The date is stored in the database in the following format
2011-10-2
If I wanted to compare today's date against the date in the database to see which one is greater, how would I do it?
I tried this,
$today = date("Y-m-d");
$expire = $row->expireDate //from db
if($today < $expireDate) { //do something; }
but it doesn't really work that way. What's another way of doing it?
If all your dates are posterior to the 1st of January of 1970, you could use something like:
$today = date("Y-m-d");
$expire = $row->expireDate; //from database
$today_time = strtotime($today);
$expire_time = strtotime($expire);
if ($expire_time < $today_time) { /* do Something */ }
If you are using PHP 5 >= 5.2.0, you could use the DateTime class:
$today_dt = new DateTime($today);
$expire_dt = new DateTime($expire);
if ($expire_dt < $today_dt) { /* Do something */ }
Or something along these lines.
in the database the date looks like this 2011-10-2
Store it in YYYY-MM-DD and then string comparison will work because '1' > '0', etc.
Just to compliment the already given answers, see the following example:
$today = new DateTime('');
$expireDate = new DateTime($row->expireDate); //from database
if($today->format("Y-m-d") < $expireDate->format("Y-m-d")) {
//do something;
}
Update:
Or simple use old-school date() function:
if(date('Y-m-d') < date('Y-m-d', strtotime($expire_date))){
//echo not yet expired!
}
I would'nt do this with PHP.
A database should know, what day is today.( use MySQL->NOW() for example ), so it will be very easy to compare within the Query and return the result, without any problems depending on the used Date-Types
SELECT IF(expireDate < NOW(),TRUE,FALSE) as isExpired FROM tableName
$today = date('Y-m-d');//Y-m-d H:i:s
$expireDate = new DateTime($row->expireDate);// From db
$date1=date_create($today);
$date2=date_create($expireDate->format('Y-m-d'));
$diff=date_diff($date1,$date2);
//echo $timeDiff;
if($diff->days >= 30){
echo "Expired.";
}else{
echo "Not expired.";
}
Here's a way on how to get the difference between two dates in minutes.
// set dates
$date_compare1= date("d-m-Y h:i:s a", strtotime($date1));
// date now
$date_compare2= date("d-m-Y h:i:s a", strtotime($date2));
// calculate the difference
$difference = strtotime($date_compare1) - strtotime($date_compare2);
$difference_in_minutes = $difference / 60;
echo $difference_in_minutes;
You can convert the dates into UNIX timestamps and compare the difference between them in seconds.
$today_date=date("Y-m-d");
$entered_date=$_POST['date'];
$dateTimestamp1 = strtotime($today_date);
$dateTimestamp2 = strtotime($entered_date);
$diff= $dateTimestamp1-$dateTimestamp2;
//echo $diff;
if ($diff<=0)
{
echo "Enter a valid date";
}
I had that problem too and I solve it by:
$today = date("Ymd");
$expire = str_replace('-', '', $row->expireDate); //from db
if(($today - $expire) > $NUMBER_OF_DAYS)
{
//do something;
}
Here's my spin on how to get the difference in days between two dates with PHP.
Note the use of '!' in the format to discard the time part of the dates, thanks to info from DateTime createFromFormat without time.
$today = DateTime::createFromFormat('!Y-m-d', date('Y-m-d'));
$wanted = DateTime::createFromFormat('!d-m-Y', $row["WANTED_DELIVERY_DATE"]);
$diff = $today->diff($wanted);
$days = $diff->days;
if (($diff->invert) != 0) $days = -1 * $days;
$overdue = (($days < 0) ? true : false);
print "<!-- (".(($days > 0) ? '+' : '').($days).") -->\n";
Found the answer on a blog and it's as simple as:
strtotime(date("Y"."-01-01")) -strtotime($newdate))/86400
And you'll get the days between the 2 dates.
This works because of PHP's string comparison logic. Simply you can check...
if ($startdate < $date) {// do something}
if ($startdate > $date) {// do something}
Both dates must be in the same format. Digits need to be zero-padded to the left and ordered from most significant to least significant. Y-m-d and Y-m-d H:i:s satisfy these conditions.
If you want a date ($date) to get expired in some interval for example a token expiration date when performing a password reset, here's how you can do:
$date = $row->expireDate;
$date->add(new DateInterval('PT24H')); // adds 24 hours
$now = new \DateTime();
if($now < $date) { /* expired after 24 hours */ }
But in your case you could do the comparison just as the following:
$today = new DateTime('Y-m-d');
$date = $row->expireDate;
if($today < $date) { /* do something */ }
first of all, try to give the format you want to the current date time of your server:
Obtain current date time
$current_date = getdate();
Separate date and time to manage them as you wish:
$current_date_only = $current_date[year].'-'.$current_date[mon].'-'.$current_date[mday];
$current_time_only = $current_date['hours'].':'.$current_date['minutes'].':'.$current_date['seconds'];
Compare it depending if you are using donly date or datetime in your DB:
$today = $current_date_only.' '.$current_time_only;
or
$today = $current_date_only;
if($today < $expireDate)
hope it helps
I have 2 dates. Lets say they look like this.
$start = 2010/12/24;
$end = 2012/01/05;
I query the database to look for visits between these two dates. I find some. I then populate an array called stats.
$stats['2010/12/25'] = 50;
$stats['2010/12/31'] = 25;
...
As you can see, there are days missing. I need to fill the missing dates with a value of zero. I was thinking something like this. (I have pulled day / month / year from start and end dates.
for($y=$start_year; $y <= $end_year; $y++) {
for($m=$start_month; $m <=$end_month; $m++) {
for($d=$start_day; $d <= $end_day; $d++) {
This would work fine for the year however the months and days wouldn't work. If the start day is the 15th. Days 1-14 of each subsequent month would be missed. I could have a solution like this then...
for($y=$start_year; $y <= $end_year; $y++) {
for($m=1; $m <13; $m++) {
$total_days = cal_days_in_month(CAL_GREGORIAN, $m, $y) + 1;
for($d=1; $d <= $total_days; $d++) {
I would then need a bunch of if statements making sure starting and end months and days are valid.
Is there a better way of doing this? Or could this even be done in my mysql query?
Just to demonstrate the power of some of PHP's newer interval handling method (mentioned by pgl in his answer):
$startDate = DateTime::createFromFormat("Y/m/d","2010/12/24",new DateTimeZone("Europe/London"));
$endDate = DateTime::createFromFormat("Y/m/d","2012/01/05",new DateTimeZone("Europe/London"));
$periodInterval = new DateInterval( "P1D" ); // 1-day, though can be more sophisticated rule
$period = new DatePeriod( $startDate, $periodInterval, $endDate );
foreach($period as $date){
echo $date->format("Y-m-d") , PHP_EOL;
}
Does require PHP >= 5.3.0
EDIT
If you need to include the actual end date, then you need to add a day to $endDate immediately before the foreach() loop:
$endDate->add( $periodInterval );
EDIT #2
$startDate = new DateTime("2010/12/24",new DateTimeZone("Europe/London"));
$endDate = new DateTime("2012/01/05",new DateTimeZone("Europe/London"));
do {
echo $startDate->format("Y-m-d") , PHP_EOL;
$startDate->modify("+1 day");
} while ($startDate <= $endDate);
For PHP 5.2.0 (or earlier if dateTime objects are enabled)
If you're using PHP5.3 then Mark Baker's answer is the one to use. If (as you say in your comment) you're still on PHP5.2 something like this should help you:
$startdate = strtotime( '2010/12/24' );
$enddate = strtotime( '2012/01/05' );
$loopdate = $startdate;
$datesArray = array();
while( $loopdate <= $enddate ) {
$datesArray[$loopdate] = 0;
$loopdate = strtotime( '+1 day', $loopdate );
}
It will create an array of the unix timestamp of every date between the start and end dates as the index and each value set to zero. You can then overwrite any actual results you have with the correct values.
$start_date = DateTime::createFromFormat('Y/m/d', '2010/12/24');
$end_date = DateTime::createFromFormat('Y/m/d', '2012/01/05');
$current_date = $start_date;
while($current_date <= $end_date) {
$current_date = $current_date->add(new DateInterval('P1D'));
// do your array work here.
}
See DateTime::add() for more information about this.
$i = 1;
while(date("Y/m/d", strtotime(date("Y/m/d", strtotime($start)) . "+ $i days")) < $end) {
... code here ...
$i++;
}
I would calculate the difference between start and end date in days, iterate on that adding a day to the timestamp on each iteration.
$start = strtotime("2010/12/24");
$end = strtotime("2012/01/05");
// start and end are seconds, so I convert it to days
$diff = ($end - $start) / 86400;
for ($i = 1; $i < $diff; $i++) {
// just multiply 86400 and add it to $start
// using strtotime('+1 day' ...) looks nice but is expensive.
// you could also have a cumulative value, but this was quicker
// to type
$date = $start + ($i * 86400);
echo date('r', $date);
}
I have this bit of horrible code saved:
while (($tmptime = strtotime('+' . (int) $d++ . ' days', strtotime($from))) && ($tmptime <= strtotime($to))) // this code makes baby jesus cry
$dates[strftime('%Y-%m-%d', $tmptime)] = 0;
(Set $from and $to to appropriate values.) It may well make you cry, too - but it sort of works.
The proper way to do it is to use DateInterval, of course.
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()))