Subtracting a certain number of hours, days, months or years from date

Question

I\'m trying to create a simple function which returns me a date with a certain number of subtracted days from now, so something like this but I dont know the date classes we

Answer 1

For hours:

function get_offset_hours($hours)
{
    return date('Y-m-d H:i:s', time() + 3600 * $hours);
}

Something like that will work well for hours and days (use 86400 for days), but for months and year it's a bit trickier...

Also you can also do it this way:

$date = strtotime(date('Y-m-d H:i:s') . ' +1 day');
$date = strtotime(date('Y-m-d H:i:s') . ' +1 week');
$date = strtotime(date('Y-m-d H:i:s') . ' +2 weeks');
$date = strtotime(date('Y-m-d H:i:s') . ' +1 month');
$date = strtotime(date('Y-m-d H:i:s') . ' +30 days');
$date = strtotime(date('Y-m-d H:i:s') . ' +1 year');

echo(date('Y-m-d H:i:s', $date));

Answer 2

something like this:

function offset hours($hours) {
    return strtotime("+$hours hours");
}

Answer 3

Try to use datetime::sub

Example from the docs (linked):

<?php

$date = new DateTime("18-July-2008 16:30:30");
echo $date->format("d-m-Y H:i:s").'<br />';

date_sub($date, new DateInterval("P5D"));
echo '<br />'.$date->format("d-m-Y").' : 5 Days';

date_sub($date, new DateInterval("P5Y5M5D"));
echo '<br />'.$date->format("d-m-Y").' : 5 Days, 5 Months, 5 Years';

date_sub($date, new DateInterval("P5YT5H"));
echo '<br />'.$date->format("d-m-Y H:i:s").' : 5 Years, 5 Hours';

?>