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03-19-2008
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changing the format of date in unix
When i execute the below command it is giving the timestamp in the format mentioned below
ls -ltr 1234.txt | awk 'BEGIN {FS=" "} {print $6" "$7" "$8}' Mar 20 00:12 i want output in the format 200803200012 please help me how to do it here year is not returned and i have to convert mar to 03 suggestions welcome thanks in advance 
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03-19-2008
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Code:
$ ls -l datediff.sh
-rw-r--r-- 1 jsaikia staff 251 Mar 10 15:06 datediff.sh
$ ls -l datediff.sh --time-style=full-iso | awk '{print $6,$7}' | awk -F ":" '{print $1,$2}' | sed -e 's/-//g' -e 's/ //g'
200803101506
//Jadu
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03-20-2008
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no the ls in my os doesnot support time-style=full-iso format
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03-20-2008
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You are going to have to use perl, python or C.
perl example - Code:
#!/bin/ksh
dfmt()
{
perl -e '
use POSIX qw(strftime);
$fmt="%Y%m%d%M%S";
$mtime=(stat $ARGV[0])[9];
($sec,$min,$hour,$day,$mon,$yr,$wday,$yday,$dntcare)=localtime($mtime);
$result=
strftime($fmt,$sec,$min,$hour,$day,$mon,$yr,$wday,$yday,$dntcare);
print "$result";
' $1
}
for file in `ls *.pl`
do
echo "$file $(dfmt $file)"
done
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