PHP datetime comparing days

Question

I'm comparing dates with this code:

$date1 = new DateTime("2007-03-24 12:10:00"); $date2 = new DateTime("2009-06-26 14:00:30"); $interval = $date1->diff($date2);

If I echo this: echo $interval->m." months and".$interval->d." days."; I get the output 3 months and 2 days.. Now, I want to echo the difference between the dates but include the amount of months in the day count, so a difference of 1 month (with 30 days in it) and 5 days would be 35 days, not 1 month and 5 days. How do I do this?

I'm using PHP version 5.3+.

@STTLCU Thanks. I didn't really think about what he wanted before posting. — Unexpected Pair of Colons
– Unexpected Pair of Colons, Commented Apr 22, 2013 at 15:15

Codeacula · Accepted Answer · 2013-04-22 15:04:52Z

5

You should be able to use:

$interval->days;

See: http://www.php.net/manual/en/class.dateinterval.php#dateinterval.props.days

echo "There are ".$interval->days." days between the two dates.";

answered Apr 22, 2013 at 15:04

Codeacula

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2 Comments

Simon Carlson Over a year ago

Well...that was easier then expected! Accepting in 9 minutes (damn timer..).

Simon Carlson Over a year ago

Quick offtopic: how do I make a datetime object from the current date and time?

STT LCU · Accepted Answer · 2013-04-22 15:06:42Z

Your $interval variable is of type DateInterval.

Therefore, $interval->days should yeld the desired output.

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