changing the format of date in unix


 
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# 1  
Old 03-20-2008
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
# 2  
Old 03-20-2008
use this way

Code:
DATE="$(date '+%Y%m%d%M%S')"
echo $DATE

# 3  
Old 03-20-2008
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

Your "ls" should support "--time-style=full-iso" for this to work.

//Jadu
# 4  
Old 03-20-2008
no the ls in my os doesnot support time-style=full-iso format
# 5  
Old 03-20-2008
Hope this works:

code:

ls -ltr data | awk 'BEGIN {FS=" "} {print $6" "$7" "$8}'|awk '$1 ~/'Feb'/{ print 200802$3}; $1 ~/Mar/{print 200803$2$3};$1 ~/Apr/{print 200804$2$3}'
# 6  
Old 03-20-2008
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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