12-01-2010, 10:09 AM
I get a date from a var and I have this to change it into unix date farmat
which is great BUT now I need to add 1 year onto that date so I can set an expiry date and hopefully later I can check against it.
How do I add 1 year onto the date? As in one year from now whenever now is.
Does that make sense.
12-01-2010, 10:16 AM
echo strtotime("+1 day"), "\n";
12-01-2010, 10:32 AM
thanks but i'm not clear
this will go to database
but I need to add on 1 year
are you saying I can do this
'".intval(strtotime($date "+365 days"))."'
12-01-2010, 10:35 AM
That depends what you are setting $date as- it has to have a value. I've never used strtotime before, so I'm just going by what I'm reading (feel free to read the link, or read the page on w3schools about strotime), but it seems like all it does is parse a date to unix format? If for $date your are using mktime or date, you can do date("y")+1 or $date = mktime(0, 0, 0, date("m"), date("d"), date("y")+1);
12-01-2010, 10:52 AM
I read / am reading the link
expected date format put into $date
00:31:02 Nov 03 2010 PDT
this '".intval(strtotime($date))."' will convert it to unix 1289049745 format
but I need to add 1 year onto that before I send to DB because I want to set an expiry date that I can check against.
thanks for trying tho, appreciate it.
if anybody can explain how I achieve this that would be great
12-01-2010, 11:34 AM
Actually, with a little reading about UNIX time format and a little testing, '".intval(strtotime($date."+1 year"))."' will return whatever you have in $date plus 1 year (31536000 seconds). If you want to add the expiration manually when you get the string, though, date("y")+1 will still work.
12-01-2010, 12:09 PM
yeh that looks like it will do the trick --- excellent effort
I was reading this page that I found
thanks very much --- :-)