How to add date + 6 month but exclude Sundays using PHP - php

Question

I am creating a scheduler script that if I executed a task today my next task will be 6 months from today (Semi-Annual) but my problem is people has no activity on Sundays so it must be adjusted on Monday instead.
How can I add 6 month to today's date but exclude Sundays, I am using PHP and able to add 6 months using below code so far.
My code is this.
<?php echo date('Y-m-d', strtotime('+6 months'));?>

Answer 1

Here's the answer: https://stackoverflow.com/a/12365635/6557808
Provides more than you're asking for and is pretty clear.

Answer 2

Here is a small function to it. Just pass from and end date
<?php
function number_of_working_days($from, $to) {
$workingDays = [1, 2, 3, 4, 5, 6]; # date format = N (1 = Monday, ...)
$from = new DateTime($from);
$to = new DateTime($to);
$to->modify('+1 day');
$interval = new DateInterval('P1D');
$periods = new DatePeriod($from, $interval, $to);
$days = 0;
foreach ($periods as $period) {
if (!in_array($period->format('N'), $workingDays)) continue;
$days++;
}
return $days;
}
echo number_of_working_days('2017-05-06', '2017-05-08');
?>

Answer 3

Complete Answer is here:--
function daysToAdd($startDate,$endDate)
{
$count=0;
While($startDate!=$endDate){
$lpcnt=0;
$begin = new DateTime($startDate);
$end = new DateTime($endDate);
$interval = DateInterval::createFromDateString('1 day');
$period = new DatePeriod($begin, $interval, $end);
foreach ( $period as $dt ){
if($dt->format( "l" )=='Sunday'){
$count++;
$lpcnt++;
}
}
$day = date('D', strtotime($endDate));
if($day=='Sun'){
$count=$count+1;
$lpcnt=$lpcnt+1;
}
$startDate=date('Y-m-d', strtotime($endDate . ' +1 day'));
$endDate=date('Y-m-d', strtotime($endDate . ' +'.$lpcnt.' day'));
if($startDate!=$endDate){
daysToAdd($startDate,$endDate);
}else{
if(date('D', strtotime($startDate))=='Sun') {
$count=$count+1;
break;
}
}
}
return $count;
}
$currentDate=date('Y-m-d');
$postSixMonth=date('Y-m-d', strtotime("+5 months"));
$daysToadd=daysToAdd($currentDate,$postSixMonth);
$requiredDate=date('Y-m-d', strtotime($postSixMonth . ' +'.$daysToadd.' days'));
echo $requiredDate;exit;

Answer 4

As stated in the PHP.net date() documentation you can check for a specific day number.
So, using that knowledge you can either:
if (date("w", strtotime("+6 months")) == 0) {
echo date("Y-m-d", strtotime("+6 months +1 days"));
}
else {
echo date("Y-m-d", "+6 months");
}
or something in the lines of Bilal Ahmed's answer using a switch
switch (date("w", strtotime("+6 months")) == 0) {
case 0: // Sunday
echo date("Y-m-d", strtotime("+6 months +1 days"));
break;
case 1: // Monday
case 2: // Tuesday
case 3: // Wednesday
case 4: // Thursday
case 5: // Friday
case 6: // Saturday
echo date("Y-m-d", "+6 months");
break;
}
which might be better if there is something to do for every different day as date() will not be executed every time you perform a case. (Note: I did not compare speeds)
EDIT: oneliner based on the comment of blokish on the askers post
echo date("w", strtotime("+6 months")) == 0 ? date("Y-m-d", "+6 months +1 days") : date("Y-m-d", "+6 months");