Shell Programming and Scripting

Hi,

I have a sample file like below:

213~!0~!Feb 16 2009 4:57:29:833PM~!0
212~!0~!Feb 7 2009 5:29:57:760PM~!0
211~!0~!Feb 4 2009 5:51:40:863PM~!0
209~!0~!Dec 17 2008 3:19:05:043PM~!0
206~!0~!Dec 4 2007 4:01:02:850PM~!0

"~!" is the field seperator.

I need to replace the first row, first column of the file with a variable passed to it and then increment the first column of the subsequent rows by 1. For eg, the variable is 500. the output should be.

500~!0~!Feb 16 2009 4:57:29:833PM~!0
501~!0~!Feb 7 2009 5:29:57:760PM~!0
502~!0~!Feb 4 2009 5:51:40:863PM~!0
503~!0~!Dec 17 2008 3:19:05:043PM~!0
504~!0~!Dec 4 2007 4:01:02:850PM~!0

Any help on this is much appreciated.

Thanks
Hemant

Try this,

cat <Filename> | sed 's/\~\!/\-/g' | awk -F'-' 'BEGIN{t=500} {print t"~\!"$2"~\!"$3"~\!"$4; t+=1}'

Hi Ramesh,

Thanks for this, it works perfect.

Just in case if this has more columns so that it is not very easy to print to print each column separatley, {print t"~!"$2"~\!"$3"~\!"$4; t+=1}. Is there a way of doing it with something like a { print $0 }, may not exactly this but something better.

Thanks
Hemant

Try this,

cat <Filename> | sed 's/\~\!/\-/g' | awk -F'-' 'BEGIN{t=500} { printf t; for(i=2; i<=NF; i++) printf "~\!"$i; printf "\n"; t+=1; }'

-srk

Thanks Ramesh, it works.

Code:

nawk -v number=500 'BEGIN{FS=OFS="~!"} {$1=number;print;number++;}' yourfile

perl:

Code:

my $num=500;
while(<DATA>){
	s/^[0-9]+/$num++/e;
	print;
}
__DATA__
213~!0~!Feb 16 2009 4:57:29:833PM~!0
212~!0~!Feb 7 2009 5:29:57:760PM~!0
211~!0~!Feb 4 2009 5:51:40:863PM~!0
209~!0~!Dec 17 2008 3:19:05:043PM~!0
206~!0~!Dec 4 2007 4:01:02:850PM~!0