Most efficient way of playing with dates?

bdm · 12-13-2007, 07:07 PM

So I've got this unix timestamp from the past which I needed to convert into a date:

PHP Code:


			
$firstDate = date('Y-m-d H:i', 1195958278);

Next I added 90 days:

PHP Code:


			
$expiry = date('Y-m-d H:i', strtotime('+90 day', $firstDate));

This is because each pass is valid for 90 days after the first date. I then proceed by figuring out how many days are left for the pass to expire:

PHP Code:


			
$daysLeft = 365 - (date('z', strtotime(date('Y-m-d H:i'))) - date('z', strtotime($expiryDate)));

I've tested it with a bit of data and it seems to work.

Does anyone see something totally wrong? Or maybe an easier alternative?

Thank you.

Village Idiot · 12-13-2007, 07:38 PM

I dont use strtotime, unix timestamps are seconds since the unix epoch (1/1/1970), so 90 days from now would be $time + (60*60*24*90), the remainder of days would be time - $original_time / (60/60/24)

Just my preference.

kevthedude · 12-13-2007, 07:57 PM

This is how I modified your code to work:

PHP Code:


			
$first = 1195958278;

$last = mktime(0, 0, 0, date("m", $first)  , date("d", $first)+90, date("Y", $first));

$today = mktime(0, 0, 0, date("m")  , date("d"), date("Y"));

$remaining = floor(($last - $today)/86400);

echo remaining;

$remaining will be your days left. You may have to switch the floor() function to ceil()... I didn't figure out the real difference in the dates but this script shows 71 days remaining. Tell me if this is wrong and I'll fix it up for you.

Salathe · 12-13-2007, 08:05 PM

As an aside, you don't need to do date('z', strtotime(date('Y-m-d H:i'))) to get today's day of the year. Simply skip the second argument to use the current time stamp: date('z').

Personally, I'd prefer working with the timestamps themselves and only use the date() function for outputting friendly strings. Something like the the following:

php Code:

<?php

$iDateStart   = 1195958278;

// expiry = starting timestamp + 90 days (in seconds)
$iDateExpire  = $iDateStart + (90 * 24 * 60 * 60);

// Days left = difference between timestamps in seconds (expiry - now)
//             divided by number of seconds in a day (86,400)
// Use floor() to round down to the day.
$iDaysLeft    = floor(($iDateExpire - time()) / (24 * 60 * 60));

// Output:
// There are 71 days left until 2008-02-23 02:37.
printf('There are %d days left until %s.',
       $iDaysLeft,
       date('Y-m-d H:i', $iDateExpire));

wGEric · 12-13-2007, 11:40 PM

Quote:

Originally Posted by Salathe

Personally, I'd prefer working with the timestamps themselves and only use the date() function for outputting friendly strings.

Quoted in agreement.

xenon · 12-14-2007, 07:46 AM

The most efficient way of working with dates in PHP it's by operating on the date timestamp and just output the formatted one (yes I am agreeing with Salathe) :D

bdm · 12-14-2007, 01:36 PM

Thanks for everyones input.

Now that I think about it, playing with the timestamp itself seems better.