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Posted by Otis on 09/04/07 07:38
PHP is driving me mad with their idiotic way of dealing with dates and
date conversions - unless I'm the idiot.
Here's what I need:
Let's assume the date is $d = "09/03/1951" - Sep 3, 1951
How do I get just the year from a date like Sep 3, 1951?
How do I get just the day from a date like Sep 3, 1951?
How do I get just the month from a date like Sep 3, 1951?
I have never been able to get PHP to deal properly with the years since
it seems to have the stupid habit of deciding for itself whether a year
is 2000 something or 1900 something. It apparently only wants to deal
with 2-digit years.
Here is what I have done in the past when I needed a FULL date:
if ($inyear >= 2000)
{
$utdatenow = strftime("%d.%m.20%y", mktime($inhours, $inmins,
$insecs, $inmonth, $inday, $inyear));
}
else
{
$utdatenow = strftime("%d.%m.19%y", mktime($inhours, $inmins,
$insecs, $inmonth, $inday, $inyear));
}
How cumbersome!!!
Isn't PHP/UNIX smart enough to know what the 4-digit year is without me
having to jump through hoops?
In Visual Basic it was simple:
yr$ = Year("09/03/1951") gave me 1951
dy$ = Day("09/03/1951") gave me 3
mo$ = Month("09/03/1951") gave me 9
Why isn't PHP as straight-forward?
Thank you.
Otis
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