Php - get date

[Nah]

In database table, there is two rows, registerDate and activeTo. Let’s say that the registerDay is ‘2/15/2020’ and activeTo is ‘2/15/2021’, is possible to actually calculate without manually writing activeTo to the database?

[John_Betong]

Yes there are umpteen different ways depending on how the dates are stored in the table and how they are used.

Beware there are different dateTime formats and “02/15/2020” and “15/02/2020” depending on the time zone:

PHP Online Manual for date(…) function

(https://www.php.net/manual/en/function.date.php

I prefer storing the table dateTime as a Unix Time Stamp because the date can be easily rendered using the PHP date(…) functin and passing the Unix time stamp:

<?php 
declare(strict_types=1);

// seconds * minutes * hours/day * days per year
   $secsPerYear = 60 * 60 * 24 * 366; 

echo '<br>$secPerYear ==> ' ,
  $secsPerYear;

echo '<br>$today ==> ' ,
  $today = date('l, jS F Y \G\M\T T - e') ;

echo '<br>$plusyear ==> ' ,
  $plusyear  = date('l, jS F Y ', time() + $secsPerYear) ;

$dd = '2/15/2020';

  $oDate = date_create($dd, timezone_open('Pacific/Nauru'));
  $oDate = date_create($dd, timezone_open('America/Chicago'));

echo '<pre>'; 
  print_r($oDate); 
echo '</pre>';

echo '<br>$newDate ==> ',
  date_format($oDate, 'Y-m-d H:i:sP') . "\n";

Output:

$secPerYear ==> 31622400
$today ==> Saturday, 15th February 2020 GMT +07 - Asia/Krasnoyarsk
$plusyear ==> Monday, 15th February 2021

DateTime Object
(
[date] => 2020-02-15 00:00:00.000000
[timezone_type] => 3
[timezone] => America/Chicago
)

$newDate ==> 2020-02-15 00:00:00-06:00

[rpkamp]

$row['registerDay'] = '2/15/2020'; // Normally this would come from the database

$registerDay = DateTimeImmutable::createFromFormat('m/d/Y', $row['registerDay']);

$activeTo = $registerDay->add(new DateInterval('P1Y')); // P1Y = 1 Year

See https://repl.it/repls/GeneralTightDisassembly for a runnable example.

[rpkamp]

@John_Betong

$secsPerYear = 60 * 60 * 24 * 366;

If you’re going this route (which I don’t recomment, the OOP way is much nicer) it should be

// date('L') = 1 when the current year is a leap year, 0 when it's not.
// Note that this assumes that the `registerDay` in the database is
// in the current year. If it's not then this calculation may well
// be incorrect.
$secsPerYear = 60 * 60 * 24 * (365 + date('L'));

UPDATE: added parentheses

[m_hutley]

You need some parenthesis there, Remon. Remember your order of operations!

[rpkamp]

Good call! I’ve added them.