PHP print dates without sunday - php

Question

I have this form with this input field
<input type="text" name="txt_datetimein" class="form-control datetime">
<input type="text" name="txt_datetimeout" class="form-control datetime">
<input type="text" name="txt_lenght" class="form-control">
I enter the first date and the second date and the length the length i precise the number of repetition
than i click next and i have all days without sunday
example if i put 5 in the length and datetimein 01-04-2017 8:00:00 and datetimeout is 01-04-2017 5:00:00
the output will be like that
Date IN Date Out Day
01-04-2017 8:00:00 01-04-2017 5:00:00 Saturday
03-04-2017 8:00:00 03-04-2017 5:00:00 Monday
04-04-2017 8:00:00 04-04-2017 5:00:00 Tuesday
05-04-2017 8:00:00 05-04-2017 5:00:00 Wednesday
06-04-2017 8:00:00 06-04-2017 5:00:00 Thursday
07-04-2017 8:00:00 07-04-2017 5:00:00 Friday
this my code but it's print all day
<?php
for($i=0;$i<=$lenght;$i++) {
$date = date('m/d/Y H:i:s', strtotime("+$i days", strtotime($datetimein)));
$edate = date('m/d/Y H:i:s', strtotime("+$i days", strtotime( $datetimeout)));
$is_sunday = date('l', strtotime($date)) == 'Sunday';
if ($is_sunday) {
$day = date('l', strtotime("+1+$i days",strtotime($datetimein)));
}
else {
$day = date('l', strtotime("+$i days",strtotime($datetimein)));
}
}
?>
How Can i solve my Problem ??!!

Answer 1

try this below code
$datetimein = "01-04-2017 8:00:00";
$datetimeout = "01-04-2017 5:00:00";
$lenght = 20;
for($i=0;$i<=$lenght;$i++) {
$date = date('m/d/Y H:i:s', strtotime("+$i days", strtotime($datetimein)));
$edate = date('m/d/Y H:i:s', strtotime("+$i days", strtotime( $datetimeout)));
$is_sunday = date('l', strtotime($date));
if($is_sunday == "Sunday")
{
$i=$i+1;
}
$day = date('l', strtotime("$i days",strtotime($datetimein)));
echo $day."<br>";
}

Answer 2

Use DateTime and all those objects. Simpler and cleaner :-)
<?php
$begin = new DateTime();
$end = clone $begin;
$end = $end->modify('+14 day');
$interval = new DateInterval('P1D');
$range = new DatePeriod($begin, $interval ,$end);
foreach($range as $date) {
if ($date->format('N') !== 7) {
echo $date->format('Y-m-d'), '<br>';
}
}
Date format N is the day of week as a number where 7 === Sunday.

Answer 3

Here is you programe
$datetimein = '01-04-2017 8:00:00';
$datetimeout= '01-04-2017 5:00:00';
$lenght = 5;
$i=0;
$days = array();
$dt = strtotime($datetimein);
while($i < $lenght){
if(date('D',$dt)!='Sun'){
$days[] = date('Y-m-d D',$dt);
$i++;
}
$dt = $dt+24*3600;
}
print_r($days);
In this line $days[] = date('Y-m-d D',$dt); change the format or save both in and out time whatever you want. $days will have you expected dates.

Answer 4

It looks like you have mis-spelled the variable length.
Check this.
<?php
$lenght = 5;
$in_temp = 0;
$out_temp = 0;
for($i=0;$i<=$lenght;$i++) {
$in = strtotime($datetimein) + $in_temp;
$out = strtotime($datetimeout) + $out_temp;
$date = date('m/d/Y H:i:s', strtotime("+$i days", $in));
$edate = date('m/d/Y H:i:s', strtotime("+$i days", $out));
$is_sunday = strtolower(date('l', strtotime($date))) == 'sunday';
if ($is_sunday) {
$in_temp += 86400 ; // Adding 1 day in seconds.
$day = date('l', (strtotime($date)+$in_temp));
}
else {
$day = date('l', (strtotime($date)));
}
echo $day."\n";
}
?>