standardizing date format

stary

Hi all,
some qns on standardizing date format.

I've this date format and value
yyyymmddhh24mi and the value, 200306121209

the date format is in oracle and the hours would have hh, hh24, hh12

thus, when splitted out, the values would be as belows
yyyy - 2003
mm - 06
dd - 12
hh24 - 12
mi - 09

how do i get the value out to this format and get this string?
2003-06-12 (yyyy-mm-dd)

your advice would be vey much appreciated.

thanks.

StaRy =)

stary

and something extra to note is that the placement of yyyy, mm, dd, hh etc etc, mi is not fixed.

the format can be displayed in any manner the user likes for example,

mmyyyyddhhmi
ddhhmi-yyyymm etc etc...

how to get the value of the string out the the format yyyy-mm-dd?

spighita

<?PHP

echo date("Y-m-d");



?>

stary

I think u've misunderstood what i'm trying to explain.
the given scenario,

value 2003061235 and format is yyyymmddhh24mi is given.

i wanted to get the value out to the string yyyy-mm-dd with the value 2003-06-12

i doesn't need today's date. what i need is to generate a dynamic string in the format yyyy-mm-dd base on the GIVEN format and value.

laserlight

Well, that should be easy enough, since it is a fixed length set format.

$given = "2003061235";
$year = substr($given, 0, 4);
$month = substr($given, 4, 2);
$day = substr($given, 6, 2);
//and perhaps
$date_str = $year . '-' . $month . '-' . $day;
//or
$date_str = substr($given, 0, 4) . '-' . substr($given, 4, 2) . '-'. substr($given, 6, 2);

stary

ermmm.. it's not a fixed string format..

user could choose to specify the following combination set
[yy] or [yyyy] for year
[dd] for day
[mm] or [mon] or [month] for month

and they can arrange in any order they like. so i think substr would not work here.

matt_4013

how about this?

$dateinput = "20031212231944";
	$format = "yyyymmddhh24MMii";
	$pos = array();
	$pos['year'] = strpos($format, "yyyy");
	$pos['month'] = strpos($format, "mm");
	$pos['day'] = strpos($format, "dd");

$year_new = substr($dateinput, $pos['year'], $pos['year'] + 3);
$month_new = substr($dateinput, $pos['month'], $pos['month'] + 1);
$day_new = substr($dateinput, $pos['day'], $pos['day'] + 1);

$date_new = "$year_new-$month_new-$day_new";