Jason,
This seems to work fine for me:
<?
$daysToAdd = 5;
$secondsToAdd = $daysToAdd * 86400;
$dateIn = "2004-07-24";
$dateInStamp = strtotime($dateIn);
echo date('m/d/Y h:i:s',$dateInStamp+$secondsToAdd);
?>... as does the more condensed version:
<? echo date('m/d/Y h:i:s',strtotime("2004-07-24")+(5*86400)); ?>If strtotime() is returning -1, that means that it's having trouble converting your YYYY-MM-DD date to a timestamp. The common pitfall is that you've got the month and day mixed up (YYYY-DD-MM), or that you're not using 2 digit numbers on the day and month (and 4 digit on the year).
If you're having trouble with strtotime(), ech out the date you're trying to convert, to make sure it's what you're expecting.
It's also worth pointing out that we've made no considerations for timezones, daylight savings, etc.
UNIX timestamps are all GMT, so if the times you're seeing are not as expected (and the dates you're collecting aren't GMT dates, then you'll need to dig a lot deeper -- it's depends how much accuracy you want.
Justin French
On 07/09/2004, at 1:59 PM, Jason FB wrote:
Can anyone tell me how to convert a date stored in the format "YYYY-MM-DD" to an integer of seconds since unix epoc,
then add $daysToAdd days to this value (I suppose if the integer was a number of seconds it would have to be $daysToAdd*60*60*24 to get the number of seconds to add)
then convert the new value to a timestamp which I can output using the function
string date ( string format [, int timestamp])
I tried doing this with mktime() and couple different ways using strtotime() and also with strftime() but I kept getting -1 as the result for strtotime() indicating the function couldn't interpret the given date.
I've been through all the date-related functions in the manual I can think of, but I'm stumpped on this one...
--- Justin French http://indent.com.au
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