Add X Days To Date, Excluding Weekends

Question

I\'m currently developing an online subscription application. I\'m having some challenges on the part where the user will select the Number of Days to subscribe and then the

Answer 1

I have developed a new function today that can help people who have this type of problem. Depends how the startdate is sent, but assuming you are using Y-m-d you can use DateTime

<?php  public static function getOrderEndDate( $start_date, $orderDaysCode ){

    $saturday_off = false;
    if( $orderDaysCode == 'meal_monthly_6' ) { $orderDays = 24; }
    elseif( $orderDaysCode == 'meal_monthly_5' ) {
        $orderDays = 20;
        $saturday_off = true;
    }elseif( $orderDaysCode == 'meal_weekly' ) {
        $orderDays = 5;
        $saturday_off = true;
    }
    else{ $orderDays = 1; }   // Daily Meal

    $formatted_date = new DateTime( $start_date );

    $date_timestamp = $formatted_date->getTimestamp();
    // loop for X days
    for( $i = 0; $i < ( $orderDays - 1 ); $i++ ) {
        // get what day it is next day
        $nextDay = date('w', strtotime('+1day', $date_timestamp) );
        // if it's Sunday or Saturday get $i-1
        if( $nextDay == 0 || ( $nextDay == 6 && $saturday_off ) ) { $i--; }
        // modify timestamp, add 1 day
        $date_timestamp = strtotime('+1day', $date_timestamp);
    }

    $formatted_date->setTimestamp($date_timestamp);

    return $formatted_date->format( 'Y-m-d' );
}

$orderEndDate = getOrderEndDate( '2020-06-17', 'meal_monthly_6' ); ?>

Answer 2

I have developed a new function today that can help people who have this type of problem.

function sumDays($days = 0, $format = 'd/m/Y') {
    $incrementing = $days > 0;
    $days         = abs($days);
    $actualDate   = date('Y-m-d');

    while ($days > 0) {
        $tsDate    = strtotime($actualDate . ' ' . ($incrementing ? '+' : '-') . ' 1 days');
        $actualDate = date('Y-m-d', $tsDate);

        if (date('N', $tsDate) < 6) {
            $days--;
        }
    }

    return date($format, strtotime($actualDate));
}

Answer 3

depends how the startdate is sent, but assuming you are using Y-m-d you can use DateTime

eg:

<?php 

    $_POST['startdate'] = '2012-08-14';
    $_POST['numberofdays'] = 10;

    $d = new DateTime( $_POST['startdate'] );
    $t = $d->getTimestamp();

    // loop for X days
    for($i=0; $i<$_POST['numberofdays']; $i++){

        // add 1 day to timestamp
        $addDay = 86400;

        // get what day it is next day
        $nextDay = date('w', ($t+$addDay));

        // if it's Saturday or Sunday get $i-1
        if($nextDay == 0 || $nextDay == 6) {
            $i--;
        }

        // modify timestamp, add 1 day
        $t = $t+$addDay;
    }

    $d->setTimestamp($t);

    echo $d->format( 'Y-m-d' ). "\n";

?>

Answer 4

Same result, shorter version:

function addDays($days,$format="Y-m-d"){

    for($i=0;$i<$days;$i++){
        $day = date('N',strtotime("+".($i+1)."day"));
        if($day>5)
            $days++;
    }
    return date($format,strtotime("+$i day"));
}