Date help

[JamesManke]

I have a variable in my database called date.

Example: Say the variable date in the databases value is 2005-05-05

How do I make this variable say...

May 5th, 2005

 

[jemminger]

http://www.php.net/manual/en/ref.datetime.php

-jeff

http://www.jeffemminger.com

try { succeed(); } catch(E) { tryAgain(); } finally { rtfm(); }
i like your sleeves...they're real big

 

[Vragabond]

Or, if you are dealing with MySQL,

http://dev.mysql.com/doc/mysql/en/date-and-time-functions.html#id2989659

 

[ronnyjljr]

Like this....

[blue]

echo date("M-d-Y", mktime(0,0,0,5,5,2005));

[/blue]

Cheers!

cout << "If you don't know where you want to go, we'll make sure you get taken";

 

[JamesManke]

I am using DATE("M-d-Y",$type[showdate])

$type[showdate] in the database is 2001-05-05

For some reason when using DATE("M-d-Y",$type[showdate]) it reads Dec-31-1969

Is this weird or what!
......heeelp!

 

[kenrbnsn]

Date is expecting a UNIX timestamp. Use:

DATE("M-d-Y",strtotime($type['showdate']))

See

http://www.php.net/strtotime

for more information.

Ken

 

[Meekos]

I'm sure this is not the best way to accomplish this but here you go.


// explode my date into an array to separate it into variables.
$mydate = explode('-',$rec['showdate']);
// separate out the components assuming format is YYYY-MM-DD
$y = $mydate[0]; // year is first
$m = $mydate[1]; // month is second
$d = $mydate[2]; // day is last

$month_name = date("F",mktime(0,0,0,$m,1,$y)); // convert to month name
$day = date("j",mktime(0,0,0,$m,$d,$y)); // remove leading zeros from the day if any
$cardinal = date("S",mktime(0,0,0,$m,$d,$y)); //get the cardinals for the day
echo "$month_name $day$cardinal, $y"; // print the date

Again, I don't know the most efficient way but this works for me.

Hope it helps.
Meekos

 

[JamesManke]

Thanks Guys! Works great - nice to a variety of options!