Date function to bypass weekends and specific dates - php

Question

I am trying to implement the code below to force dates to skip weekends and holiday array. The weekend part is working but I am having difficulty with the holiday section, here is my code:
$holidayDates = array(
'10-10-2017', '11-10-2017'
);
$fcount = 0;
$temp = strtotime("9-10-2017");
while ($fcount < 1) {
$nextfcount = strtotime('+1 weekday', $temp);
$nextfcount2 = date('d-m-Y', $temp);
if (!in_array($nextfcount2, $holidayDates)) {
$fcount++;
}
$temp = $nextfcount;
}
$newDater = date("d-m-Y", $temp);

Answer 1

Let me debug this for you:
$holidayDates = array(
'10-10-2017', '11-10-2017'
);
$fcount = 0;
// 1. $temp is a terrible name for a variable
$temp = strtotime("9-10-2017");
// 2. in other words - if $fcount == 1, exit loop
while ($fcount < 1) {
// 3. $temp is now the 9th
$nextfcount = strtotime('+1 weekday', $temp);
// 4. $nextfcount2 is now the 10th
$nextfcount2 = date('d-m-Y', $temp);
if (!in_array($nextfcount2, $holidayDates)) {
// 5. Yes, found your date. $fcount is now 1
$fcount++;
}
// 6. $temp is now the 10th
$temp = $nextfcount;
// 7. $fcount is now 1, `while` condition is met, exit loop
}
// 8. The date is now the 10th
$newDater = date("d-m-Y", $temp);
You question is much too vague for me to provide you with a quality answer.
I will assume, you want to get 12-10-2017 as the result of your loop. This code will do the trick:
<?php
$holidayDates = [
'10-10-2017',
'11-10-2017',
];
$foundCounter = 0;
$theDate = strtotime("9-10-2017");
while ($foundCounter < 1) {
$theDate = strtotime('+1 weekday', $theDate);
if (!in_array(date('d-m-Y', $theDate), $holidayDates)) {
$foundCounter++;
}
}
$newDater = date("d-m-Y", $theDate);
var_dump($newDater);
// string(10) "12-10-2017"
https://3v4l.org/DcWBZ