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Posted by Steve on 09/04/07 13:31
> unless I'm the idiot.
which is totally the case. don't bash php because *you* don't get it. rtfm!
> Here's what I need:
>
> Let's assume the date is $d = "09/03/1951" - Sep 3, 1951
$d = strtotime('09/03/1951'); // which won't work on windows server
> How do I get just the year from a date like Sep 3, 1951?
echo date('Y', $d);
> How do I get just the day from a date like Sep 3, 1951?
echo date('d', $d);
> How do I get just the month from a date like Sep 3, 1951?
echo date('m');
really hard, ain't it?!!!
> I have never been able to get PHP to deal properly with the years since it
> seems to have the stupid habit of deciding for itself whether a year is
> 2000 something or 1900 something. It apparently only wants to deal with
> 2-digit years.
ALL software languages have to ASSUME what year you want when you give it a
two-digit year, moron! AND MOST ALL OF THEM use 38 as the magic number. if
below or eq 38, 20 + two-digit is used...if > 38, 19 + two-digit. RTFM !!!
> Here is what I have done in the past when I needed a FULL date:
>
> if ($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, $inyear));
> }
>
> How cumbersome!!!
yep...never mind it being stupid.
> Isn't PHP/UNIX smart enough to know what the 4-digit year is without me
> having to jump through hoops?
lol...aren't you smart enough to read a fucking manual?
> In Visual Basic it was simple:
ahhhh...NOW THAT EXPLAINS IT!!!
> yr$ = 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?
it is, AND it is *easier* because you only need ONE function in the first
place in order to get the result(s) you want!
> Thank you.
piss off.
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