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

Otis schreef:
> 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?
>

Use the calendar extension:

http://www.php.net/calendar


JW

These don't me with the problems I described.


otis



Janwillem Borleffs wrote:
> Otis schreef:
>> 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?
>>
>
> Use the calendar extension:
>
> http://www.php.net/calendar
>
>
> JW

Otis schreef:
> These don't me with the problems I described.
>

$d = 'Sep 3, 1951';

$y = strftime('%Y', strtotime($d));
$m = strftime('%m', strtotime($d));
$d = strftime('%d', strtotime($d));


JW

..oO(Otis)

>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?

Internally PHP works with Unix timestamps. Use strtotime() to turn your
date into such a timestamp, then use whatever date function you like,
e.g.

$date = getdate(strtotime('09/03/1951'));

Now $date will be an array with all the informations you want:

Array
(
[seconds] => 0
[minutes] => 0
[hours] => 0
[mday] => 3
[wday] => 1
[mon] => 9
[year] => 1951
[yday] => 245
[weekday] => Monday
[month] => September
[0] => -578451600
)

Micha

> 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.

On Tue, 04 Sep 2007 00:38:22 -0700, Otis wrote...
>
>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


Might check either of there at the PHP site. I believe the top one will give you
a string of text with the information you request. The second would give you an
array you can probably get more use out of...

http://www.php.net/manual/en/function.date.php
http://www.php.net/manual/en/function.getdate.php

Tom
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"Otis" <otie_nospam@cox.net> wrote in message
news:r5aDi.86468$Mu5.80898@newsfe15.phx...
> These don't me with the problems I described.
>
>
> otis

dude! pull your head out of your ass!!!