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Top Forums UNIX for Dummies Questions & Answers awk - Print lines if only matching key is found Post 302933978 by High-T on Tuesday 3rd of February 2015 04:28:17 PM
Old 02-03-2015
I have updated the question with the script.

Code:
awk 'FNR == NR && ! /^[[:space:]]*$/ { key = substr($0,1,8); a[key] = $0; next }
$1 == "01" { if (key != 0)
             {
                 if (key in a)
                 {
                     printf("77\t%s\n", a[key])
                     delete a[key]
                 }
             }
             key = $4 $3 $2
           }
{ if (key in a) print $0 } ' \
     File2.txt File1.txt > File3.txt

 

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GREP(1) 						      General Commands Manual							   GREP(1)

NAME
grep - search a file for lines containing a given pattern SYNOPSIS
grep [-elnsv] pattern [file] ... OPTIONS
-e -e pattern is the same as pattern -c Print a count of lines matched -i Ignore case -l Print file names, no lines -n Print line numbers -s Status only, no printed output -v Select lines that do not match EXAMPLES
grep mouse file # Find lines in file containing mouse grep [0-9] file # Print lines containing a digit DESCRIPTION
Grep searches one or more files (by default, stdin) and selects out all the lines that match the pattern. All the regular expressions accepted by ed and mined are allowed. In addition, + can be used instead of * to mean 1 or more occurrences, ? can be used to mean 0 or 1 occurrences, and | can be used between two regular expressions to mean either one of them. Parentheses can be used for grouping. If a match is found, exit status 0 is returned. If no match is found, exit status 1 is returned. If an error is detected, exit status 2 is returned. SEE ALSO
cgrep(1), fgrep(1), sed(1), awk(9). GREP(1)
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