Confused about date-time difference in months and years - php

Question

I've made a function to calculate the age between two date-times, but I had a problem with the months and the years, because the months could be 29, 30, or 31 days.
I am confused, is this going to cause any problem? and if yes, then could someone suggest a solution please.
HERE IS MY FUNCTION
function time_age($date_time1,$date_time2 = ''){
//store current date-time if it hasn't been given
if (empty($date_time2)){
$date_time2 = date('Y-m-d H:i:s');
}
//get date-time difference in seconds
$time_age = strtotime($date_time2) - strtotime($date_time1);
//to store if seconds, minutes, hours, days, weeks, months, years
$time_age_type = 'seconds';
if ($time_age >= 60 and $time_age < 3600){
$time_age_type = 'minutes';
$time_age = number_format($time_age / 60, 0);
}elseif ($time_age >= 3600 and $time_age < 86400){
$time_age_type = 'hours';
$time_age = number_format($time_age / 3600, 0);
}elseif ($time_age >= 86400 and $time_age < 604800){
$time_age_type = 'days';
$time_age = number_format($time_age / 86400, 0);
}elseif ($time_age >= 604800 and $time_age < 2629743){
$time_age_type = 'weeks';
$time_age = number_format($time_age / 604800, 0);
}elseif ($time_age >= 2629743 and $time_age < 31556926){
$time_age_type = 'months';
$time_age = number_format($time_age / 2629743, 0);
}elseif ($time_age >= 31556926){
$time_age_type = 'years';
$time_age = number_format($time_age / 31556926, 0);
}
return $time_age.' '.$time_age_type.' ago';
}

Answer 1

I will suggest this alternative solution so I took this example from https://php.net :
<?php
$val1 = '2014-03-18 10:34:09.939';
$val2 = 'now';
$datetime1 = new DateTime($val1);
$datetime2 = new DateTime($val2);
echo "<pre>";
var_dump($datetime1->diff($datetime2));
https://secure.php.net/manual/en/datetime.diff.php#114656

Answer 2

How about this function which is what I use?
/**
* Get a string of length of time elapsed since a specified time.
*
* #param int $time the specified time to check
* #return string returns a string stating the length of time ago.
*/
function timeSince($time)
{
$timeDifference = time() - $time;
$timeTokens = [
31536000 => 'year',
2592000 => 'month',
604800 => 'week',
86400 => 'day',
3600 => 'hour',
60 => 'min',
1 => 'sec'
];
foreach ($timeTokens as $unit => $text) {
if ($timeDifference < $unit) continue;
$noUnits = floor($timeDifference / $unit);
return $noUnits . " " . $text . (($noUnits > 1) ? "s ago" : " ago");
}
}
Usage
echo timeSince(time() - 86400);
Output
1 day ago
You can also use the DateTime class' diff function:
An OO approach:
$datetime1 = new DateTime('2009-10-11');
$datetime2 = new DateTime('2009-10-13');
$interval = $datetime1->diff($datetime2);
echo $interval->format('%R%a days');
A Procedural approach:
$datetime1 = date_create('2009-10-11');
$datetime2 = date_create('2009-10-13');
$interval = date_diff($datetime1, $datetime2);
echo $interval->format('%R%a days');
Note: If you are using the DateTime class and wish to find the difference between two UNIX timestamps then you have to put an # sign before the timestamp, for example:
new DateTime("#" . $yourUnixTimestamp);
Otherwise it will throw an error.