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Parsing a date with PHP

Of course we all know we can get the system date from:

date("m-d-Y H:i:s");

However....assuming that I have a variable called $mydate with this data already in it, I need to seperate the data back out into its companents.

For example if we assume

$mydate equals "2003-11-09 12:56:19"

I would like now to parse out that variable so I can get the year, month, day, hour, minute and second back.

thanks

Michael701

<?PHP
$mydate="2003-11-09 12:56:19";

echo 'strtotime()='.strtotime($mydate)." \n";

$mydate_array = getdate(strtotime($mydate));

echo 'Year='.$mydate_array['year']." \n";
echo 'Month='.$mydate_array['mon']." \n";
echo 'Day='.$mydate_array['mday']." \n";
?>

outputs

strtotime()=1068404179
Year=2003
Month=11
Day=9

AlanJDM

or...

<?PHP
$mydate="2003-11-09 12:56:19";
echo 'Year='.date("Y",strtotime($mydate))." \n";
echo 'Month='.date("m",strtotime($mydate))." \n";
echo 'Day='.date("d",strtotime($mydate))." \n";
echo 'Hour='.date("g",strtotime($mydate))." \n";
echo 'Month='.date("i",strtotime($mydate))." \n";
echo 'Second='.date("s",strtotime($mydate))." \n";
?>

outputs

Year=2003
Month=11
Day=09
Hour=12
Month=56
Second=19

Just another way to do it, does the same thing that Michael701 did.

Alan

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yahelb

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webcs

ASKER

Great answers...

Note to self, make sure to replace the g: with an H: when doing the parsing to get military time and not go nutty.

Looked for an hour for an error that wasn't there...figures :)