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localtime format as "1999-01-28"

Posted on 1999-01-28
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Last Modified: 2010-03-05
Hi.......
I am using ....
$DATE = (join'-',(split/ /,localtime)[4,1,2]);
this returns "1999-Jan-28", but I need the Month to be a digit. "1999-01-28".
regards
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Question by:rickyr
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9 Comments
 
LVL 4

Expert Comment

by:martinag
ID: 1210350
@time = localtime;
$DATE = sprintf "%d-%02d-%02d", $time[5]+1900, $time[4]+1, $time[3];

Martin
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LVL 4

Expert Comment

by:martinag
ID: 1210351
Or to make it a one-liner...
$DATE = sprintf "%d-%02d-%02d", (localtime)[5]+1900, (localtime)[4]+1, (localtime)[3];

Martin
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LVL 84

Expert Comment

by:ozo
ID: 1210352
#also:
use POSIX;
$DATE = strftime("%Y-%m-%d",localtime);
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LVL 84

Expert Comment

by:ozo
ID: 1210353
#another one-liner:
$DATE = sprintf"%d-%02d-%02d",map{$_+(0,1,1900)[-(?^?..!??)]}(localtime)[5,4,3];
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LVL 3

Author Comment

by:rickyr
ID: 1210354
hi....
I've been using the POSIX one, mainly coz I wanted to add %H-%M-%S for the tme aswell.
thanx, please answer.
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LVL 84

Expert Comment

by:ozo
ID: 1210355
$time=sprintf"%02d-%02d-%02d",(localtime)[2,1,0];
#or
use POSIX;
$time=strftime("%H-%M-%S",localtime);
#or
$time = join'-',(split/\W/,localtime)[3..5];
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LVL 3

Author Comment

by:rickyr
ID: 1210356
Thanx ozo, but I meant re-post as an answer.

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LVL 3

Author Comment

by:rickyr
ID: 1210357
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LVL 84

Accepted Solution

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ozo earned 10 total points
ID: 1210358
answer
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