Date problem in PHP - php

Question

I have a piece of PHP that is trying to do this:
1) given a string like "h m s" (where h=hr, m=min, s=sec)
2) Add the time from 1) to time()
3) format the result to look like "y-mth-d-h-min-s"
So say the time is now 01-01-2011 1am, I want it to add "10 0 0", which should give me 01-01-2011 11am, but for some reason at the moment, it does seem to add the string, but it's not accurate.
This is the code I'm using:
$values_arr['regx_expdate'] = date("Y-m-d H:i:s", time()+$values_arr['regx_expdate']);
where $values_arr['regx_expdate'] is the string in the format "h m s", eg. "10 0 0".
The main question is how would time() know if "10 0 0" is actually 10hrs 0min 0min, and not 10days 0hr 0min??

Answer 1

It does not.
It will cast it to int, interpret it as seconds and add it to the result of time().
Some code that could do as you describe would be:
list ($h,$m,$s) = explode(' ', $values_arr['regx_expdate'], 3);
$difference = 60*60*$h + 60*$m + $s;
$values_arr['regx_expdate'] = date("Y-m-d H:i:s", time()+$difference);

Answer 2

Easiest method I can think of is to extract each token from $values_arr['regx_expdate'], add up the seconds and simple add it to time().
For example
if (preg_match('/^(\d{1,2}) (\d{1,2}) (\d{1,2})$/', $values_arr['regx_expdate'], $units)) {
$seconds = $units[3] + ($units[2] * 60) + ($units[1] * 3600);
$newTimestamp = time() + $seconds;
}

Answer 3

After parsing your input string into Hours Minutes and Seconds, it might be worth reorganizing said array of values into a string that PHP's strtotime can process.

Answer 4

This little function may help, you may customize it to suit your purpose:
function AddingDaysFromNow($number_of_days)
{
$today = mktime(0, 0, 0, date('m'), date('d'), date('Y'));
// today is now time return in seconds
$addingTime = $today + (86400 * $number_of_days);
// adding to advance it
//choice a date form at here
return date("Y-m-d", $addingTime);
}
//use it as
$expireDate = AddingDaysFromNow(2); // assume the 2 is advance to 2 days ahead
// return the day in future
Good luck!

Answer 5

To handle and convert dates in php you should first force everything into unixtimestamp and then you give it the structure you want
$date = date("THE-DATE-FORMAT-YOU-WANT", "THE-DATE-YOU-WOULD-LIKE-TO-CONVERT-IN-SECONDS");
//For example.
$new_date = date("Y-m-d H:i:s", strtotime($old_date));
$now = date("Y-m-d H:i:s", time());