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convert date format

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# 1  
Old 04-28-2011
convert date format
I've been using this thread:
https://www.unix.com/shell-programmin...m-dd-yyyy.html
and
https://www.unix.com/shell-programmin...dd-ddmmyy.html
and this code:
Quote:
awk 'BEGIN {FS="/";OFS="/"}{print $3 $1 $2 }' infile > outfile
on this format:
Code:
05/16/2008 18:30:49  Installation  48985

and I'm real close, but I got this:
Code:
2008 18:30:49  Installation  489851605

So, I'm not getting a col. setting correct, It puts $1, $2 at the end of the line and my / separator is missing.

help appreciated.
# 2  
Old 04-28-2011
Code:
awk -F '[ /]'  '{print $1, $2, $3}'

you need to have more than one FS value. awk accepts a character class [ ... ]
specitifcation. Solaris - use nawk.
# 3  
Old 04-28-2011
Code:
echo '05/16/2008 18:30:49  Installation  48985' | nawk '{split($1,a,"/");$1=a[3] "/" a[2] "/" a[1]}1'
echo '05/16/2008 18:30:49  Installation  48985' | sed 's#^\(..\)/\(..\)/\(....\)#\3/\2/\1#'

Last edited by vgersh99; 04-28-2011 at 02:40 PM..
This User Gave Thanks to vgersh99 For This Post:
dba_frog
# 4  
Old 04-29-2011
using perl
Code:
 echo '05/16/2008 18:30:49  Installation  48985' | perl -pe 's/(\d{2})\/(\d{2})\/(\d{4})(.*)/$3\/$2\/$1$4/'

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