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Changing the date format

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# 1  
Old 03-01-2012
Changing the date format
Hi all,

I have a file with below data
Code:
af23b|11-FEB-12|acc7
ad23b|12-JAN-12|acc4
as23b|15-DEC-11|acc5
z123b|18-FEB-12|acc1

I need the output as below:-(date in yyyymmdd format)
Code:
af23b|20120211|acc7
ad23b|20120112|acc4
as23b|20111215|acc5
z123b|20120218|acc1

Please help me on this.
Thanks in adv.


Moderator's Comments:
Mod Comment How to use code tags
Last edited by Franklin52; 03-01-2012 at 03:10 AM.. Reason: Please use code tags for code and data samples, thank you
# 2  
Old 03-01-2012
If you have GNU date, try this:
Code:
while IFS='|' read a b c; do echo "$a|`date -d$b +%Y%m%d`|$c"; done < inputfile

This User Gave Thanks to balajesuri For This Post:
gani_85
# 3  
Old 03-01-2012
Dear Balajesuri,

Thanks for your reply. I have used the below command, but it is not working , i have korn shell, please help me for this..
Code:
awk '{while IFS='|' read a b c; do echo "$a|`date -d$b +%Y%m%d`|$c"; done}' < inputfile


Or , please give me a command which will convert the date 11-JAN-12 to 20120111

thanks
Last edited by Franklin52; 03-01-2012 at 03:10 AM.. Reason: Please use code tags for code and data samples, thank you
# 4  
Old 03-01-2012
Quote:
Originally Posted by gani_85
\i have korn shell, please help me for this..

awk '{while IFS='|' read a b c; do echo "$a|`date -d$b +%Y%m%d`|$c"; done}' < inputfile
It's not working because you put that thing in an awk statement. Just put it as is on command line (ofcourse by substituting the filename to whatever contains your input)

Also, check if you have GNU date using this command: date --version
This User Gave Thanks to balajesuri For This Post:
gani_85
# 5  
Old 03-01-2012
Dear Balajesuri,

No, i dont have GNU date.
I am working in AIX system
# 6  
Old 03-01-2012
Code:
#! /usr/bin/perl -w
use strict;

open I, "< inputfile";
for (<I>) {
    chomp;
    my @x = split /\|/;
    $x[1] = parse_date ($x[1]);
    print join ('|', @x), "\n";
}
close I;

sub parse_date {
    my $t = shift;
    my @d = split /-/, $t;
    my %mnths = (   "JAN" => "01", "FEB" => "02", "MAR" => "03", "APR" => "04", "MAY" => "05", "JUN" => "06",
                    "JUL" => "07", "AUG" => "08", "SEP" => "09", "OCT" => "10", "NOV" => "11", "DEC" => "12" );
    for (keys %mnths) { if ($d[1] eq $_) { $d[1] = $mnths{$_}; last } }
    return "20$d[2]$d[1]$d[0]";
}

I've hard-coded it to print "20" for the first 2 digits of year assuming you won't have to deal with the last or next century for now, atleast.

And, please start your posts by mentioning which system and shell you're working on. It helps in getting quicker answers.
This User Gave Thanks to balajesuri For This Post:
gani_85
# 7  
Old 03-01-2012
I know its dirty but worth to post.

Code:
$ mth=JANFEBMARAPRMAYJUNJULAUGSEPOCTNOVDEC
$ awk -F\| -v mlist=$mth '{split($2,A,"-") ;printf("%s|20%d%02d%d|%s\n", $1,A[3],(index(mlist,A[2])+2)/3,A[1],$3)}' x
af23b|20120211|acc7
ad23b|20120112|acc4
as23b|20111215|acc5
z123b|20120218|acc1
$


You having all caps in month name. Therefore my string looks dirty.
With standard 3 letters name, its looks "JanFebMarAprMayJunJulAugSepOctNovDec". Somewhat meaningful.
This User Gave Thanks to clx For This Post:
gani_85
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