CODEWITHSUNDEEP
bashunixawk
Bhaskar Mishra
Bhaskar Mishra

Reputation: 3542

Awk command to format text output with custom record separator and field Separator

I have a file which has the data in the following manner : 

1
2
3
end
4
5
6

I want an output in the form : 

1   2   3
4   5   6

here the string "end" represents the record separator. I am new to UNIX and used Awk to achieve this, I am sure there can be much easier and a better way. Here is my attempt: 

awk 'BEGIN {RS="end\n"; FS="\n"} { print $1 "\t"  $2 "\t" $3 "\t" $4'} atestfile.awk

Upvotes: 2

Views: 535

Answers (2)

Sobrique
Sobrique

Reputation: 53478

I'd probably use perl (and yes, I know that's not quite what you asked, but given it's standard on Unix, I'll offer it anyway).

use strict;
use warnings;

#read STDIN or a filename designated on command line. 
while (<>) { 
    #replace linefeeds with spaces. 
    s/\n/ /g;
    #Check for presence of "end"
    if (m/end/) {
        #insert linefeed
        print "\n";
    }
    else {
        print;
    }
}

Upvotes: 0

Tom Fenech
Tom Fenech

Reputation: 74615

Personally I'd write it like this:

awk -v RS='end' -v OFS='\t' '{$1=$1}1' file

The differences are that I've used -v to specify the record separator RS and the output field separator OFS. I've also used $1=$1 to make awk "touch" each record (so that the format is changed to use OFS) and 1 at the end as a shorthand for {print} (since 1 is always True and the default action of awk on True is to print the record).

Upvotes: 6

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