I have a cross reference file which contains 86000 records. The data is old number:new number. There are 100s of files where i need to search for old number and append corresponding new number (preceded by @) to the line containing old number. The files contain millions of records.
Currently I am using sed command as below:
v_s_replace_string = contains old number
and v_s_new_string = contains new number
I need to replace this sed command with and awk command as awk is faster than sed.
Please help.
Thanks
Last edited by Franklin52; 10-05-2009 at 03:35 AM..
Reason: Please use code tags!
How do I alter this command so that it prints only the second comma delimited field from line number 3? Secondly, how do you redirect the output to a variable called TEST?
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Discussion started by: maddyfreaks1
3 Replies
LEARN ABOUT LINUX
igawk
IGAWK(1) Utility Commands IGAWK(1)NAME
igawk - gawk with include files
SYNOPSIS
igawk [ all gawk options ] -f program-file [ -- ] file ...
igawk [ all gawk options ] [ -- ] program-text file ...
DESCRIPTION
Igawk is a simple shell script that adds the ability to have ``include files'' to gawk(1).
AWK programs for igawk are the same as for gawk, except that, in addition, you may have lines like
@include getopt.awk
in your program to include the file getopt.awk from either the current directory or one of the other directories in the search path.
OPTIONS
See gawk(1) for a full description of the AWK language and the options that gawk supports.
EXAMPLES
cat << EOF > test.awk
@include getopt.awk
BEGIN {
while (getopt(ARGC, ARGV, "am:q") != -1)
...
}
EOF
igawk -f test.awk
SEE ALSO gawk(1)
Effective AWK Programming, Edition 1.0, published by the Free Software Foundation, 1995.
AUTHOR
Arnold Robbins (arnold@skeeve.com).
Free Software Foundation Nov 3 1999 IGAWK(1)