Adding years to a date resets to 1970-01-01

Question

$somedate = \"1980-02-15\";
$otherdate = strtotime(\'+1 year\', strtotime($somedate));
echo date(\'Y-m-d\', $otherdate);

outputs

19         


        

Answer 1

PHP's dates are limited to a range from 01-01-1970 to 19-01-2038. You will have to use a different method for working with dates.

PEAR has a Date class : PEAR Date

Answer 2

strtotime() uses a unix timestamp, so it overflows if it attempts to calculate years beyond 2038 and reverts back to 1970.

To get around this, use the DateTime object. http://php.net/manual/en/book.datetime.php

To add a period of time to a DateTime object, use DateTime::add, which takes a DateInterval as a parameter. http://php.net/manual/en/datetime.add.php http://www.php.net/manual/en/class.dateinterval.php

$date = new DateTime("1980-02-15");
if (method_exists("DateTime", "add")) {
    $date->add(new DateInterval("Y75"));
} else {
    $date->modify("+75 years");
}
echo $date->format("Y-m-d");

Answer 3

It's the 2038 bug which is like y2k where systems can't handle dates after that year due to 32 bit limitations. Use the DateTime class instead which does work around this issue.

For PHP 5.3+

$date = new DateTime('1980-02-15');
$date->add(new DateInterval('P75Y'));
echo $date->format('Y-m-d');

For PHP 5.2

$date = new DateTime('1980-02-15');
$date->modify('+75 year');
echo $date->format('Y-m-d');

Answer 4

For unix timestamp, the maximum representable time is 2038-01-19. At 03:14:07 UTC.

So you can't represent/operate time that over it by using timestamp.

Answer 5

75 years from 1980 is 2055, which is past the highest date value that can be represented in a 32-bit integer. Therefore the result becomes 0, which is the 1970 date you observe.