this is an option that only "sort of" works (converting to date/time);
$x = strtotime("2002-04-03") - strtotime("2002-04-02");
$x will contain 86400 which is the number of seconds in a "normal" day.
Since the days that some people use switch from daylight savings time to
standard time (and visa versa) there are a different number of seconds in
those days so simply dividing by some fixed value will not work (this is
what I mean by "sort of").
Some databases have a datadiff function, perhaps you can pass a query
something like
SELECT datediff("2002-04-03","2002-04-02") as difference from realtable
Warren Vail
Tools, Metrics & Quality Processes
(415) 667-7814
Pager (877) 774-9891
215 Fremont 02-658
-----Original Message-----
From: Rick Emery [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, April 03, 2002 11:02 AM
To: 'ROBERT MCPEAK'; [EMAIL PROTECTED]
Subject: RE: [PHP] syntax for date math expressions
Convert to date/time variable and perform arithmetic.
Otherwise, if these dates are from mysql, let mysql do it
-----Original Message-----
From: ROBERT MCPEAK [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, April 03, 2002 11:12 AM
To: [EMAIL PROTECTED]
Subject: [PHP] syntax for date math expressions
What is the php syntax for adding or subtracting dates?
For example, I'd like to do something like this:
2002-04-03 - 2002-04-02 = 1
or
2002-04-03 - 2002-04-02 = 0000-00-01
or
2002-04-03 + 0000-00-01 = 2002-04-03
etc.
Can anybody help this newbie?
Thanks!
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