Date conversion

[Mike_Ca Li]

HI :
I have eg:
01Jan04
24Jun04
how to convert to
01-Jan-04
24-Jun-04 ?
Thanks

[Michael Tully]

Try this:

date %d%b%y

[Patrick Wallek]

#!/usr/bin/sh
DATE=$1
DAY=$(echo ${DATE} | cut -c 1-2)
MONTH=$(echo ${DATE} | cut -c 3-5)
YEAR=$(echo ${DATE} | cut -c 6-7)

echo "${DAY}-${MONTH}-${YEAR}"


Invoke it as:

# conv_date 01Jan04
And the output should be:
01-Jan-04

[Michael Schulte zur Sur · Accepted solution]

Hi,

like
echo "01Jan04" | awk '{print substr($1,1,2)"-"substr($1,3,3)"-"substr($1,6,2)}'

hth,

Michael

[Michael Tully]

obviously I didn't read the question correctly ... sorry

[Rodney Hills]

How about this perl one liner-

perl -p -e 's/(..)(...)(..)/$1-$2-$3/' inpfile >outfile

HTH

-- Rod Hills

[Hein van den Heuvel]

perl -pe 's/(\D+)/-$1-/' infile > outfile

Now this of course will only work with the very simple examples you show.

Just for grins,

Hein.

[H.Merijn Brand (procura]

And for *all* other formats (including those of your example):

# perl -wMDate::Manip -e'$dt=UnixDate($string,"%d-%b-%Y");

Enjoy, Have FUN! H.Merijn

[H.Merijn Brand (procura]

And for *all* other formats (including those of your example):

# perl -wMDate::Manip -e'$dt=UnixDate($string,"%d-%b-%Y")'

where $string is your date
if the dates are in a file

# perl -wMDate::Manip -pe'$_=UnixDate($_,"%d-%b-%Y")' file

Enjoy, Have FUN! H.Merijn

[Saravanan_11]

hai ,

Use in the following form date "+%d-%h-%y"

# date "+%d-%h-%y"
29-Jul-04
#

-Saravanan