Countdown time until end date

Question

I am trying to report how many hours are remaining of a users subscription. The format of date/time which I am using is Y-m-d H:i:s. Basically I log the time the user pays in the database, and assume the plan will expire 1 day from when the payment was made.

What is wrong with my logic here in the calculation of difference between the current time and the end date?

    $start_time = strtotime($callback->_time_paid());
    $end_time   = 0;

    $end_time = strtotime(date('Y-m-d H:i:s', strtotime($callback->_time_paid().(' +'.$days. 'day'))));

    $static = ($end_time - $start_time);
    $time = (strtotime(date('Y-m-d H:i:s')) - $static) - $static;
    print_r($time);

    $hours = floor($time / 3600);
    $minutes = floor($time / 60);

Possible duplicate of [Calculate number of hours between 2 dates in PHP](https://stackoverflow.com/questions/3108591/calculate-number-of-hours-between-2-dates-in-php) — fubar, Oct 26 '18 at 00:47
almost all uses of strtotime should be replaced with: http://php.net/manual/en/class.datetime.php — , Oct 26 '18 at 00:52

score 0 · Answer 1 · answered Oct 26 '18 at 01:41

Not sure why you are subtracting the difference in seconds twice, or the math on your conversions but I've not had enough coffee yet.

Here's what works for me - get your timestamp from DB, if it is in the future then do some math.

<?php

$end_time = strtotime(date('Y-m-d H:i:s', strtotime($callback->_time_paid().(' +'.$days. 'day'))));

if($end_time>(time())){
    // how many seconds are we dealing with?
    $seconds=$end_time-(time());
    $hours=(int)($seconds/3600);
    $min=(int)(($seconds%3600)/60);
    $sec=(($seconds%3600)%60);
    print($hours." ".$min." ".$sec);
}else{
    print("Expired already");
}

?>

Sorry for the over abundant parentheses, but I have math issues and I force myself to always be extremely explicit with it even though I know the order of operations — ivanivan, Oct 26 '18 at 01:42

Countdown time until end date

1 Answers1