How to display today and the next four days php - arrays

Question

I have an app where I need to display today, and the next four days on.
$daysOn = [ 'Mon', 'Tue', 'Wed', 'Thu', 'Fri', 'Mon'];
$daysOff = [ 'Sat', 'Sun' ];
$today = date("N", strtotime('today'));
$yesterday = date("N", strtotime('yesterday'));
$daysOnRearranged = array_merge(array_slice($daysOn, $today), array_slice($daysOn, 0, $yesterday));
This is currently showing me:
Monday, Monday, Tuesday, Wednesday, Thursday.
I need to show today, and the next for days on.
Any ideas?

Answer 1

Here I am using the strtotime ability to workout the date using a string like "Today + 1 day". I am also using the character "D" that returns 'A textual representation of a day, three letters'. You can change the "D" to a "l" if you want the full name of the day.
$nextDays = [];
for ($i = 0; $i <= 4; $i++) {
$nextDays[] = date("D", strtotime("today + $i day"));
}
var_dump($nextDays);
To remove weekends:
$nextDays = [];
$daysOff = ["Sat", "Sun"];
$n = 0;
do {
$day = date("D", strtotime("today + $n day"));
if (!in_array($day, $daysOff)) { // Check if the above $day is _not_ in our days off
$nextDays[] = $day;
}
$n ++; // n is just a counter
} while (count($nextDays) <= 4); // When we have four days, exit.
var_dump($nextDays);

Answer 2

On days check condition to be checked inside for loop.
Please have a look into below snippet and running code.
<?php
function pr($arr = [])
{
echo "<pre>";
print_r($arr);
echo "</pre>";
}
$daysOn = ['Mon', 'Tue', 'Wed', 'Thu', 'Fri', 'Mon'];
$daysOff = ['Sat', 'Sun'];
$today = date("N", strtotime('today'));
$nextDays = [];
$j = $i = 0;
while($j != 4){ // until it found next 4 working days
$day = date('D', strtotime("today +$i day")); // it will keep checking for next working days
if(in_array($day, $daysOn)){ // will check if incremental day is in on days or not
$nextDays[] = $day; // it will add on days in result array
$j++;
}
$i++;
}
pr($nextDays);die;
Here is working demo.
EDIT
Above snippet is for checking on days.
You can check for OFF Days like below.
while($j != 4){
$day = date('D', strtotime("today +$i day"));
if(!in_array($day, $daysOff)){
$nextDays[] = $day;
$j++;
}
$i++;
}
pr($nextDays);die;