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Top Forums Shell Programming and Scripting can I pass awk variable to system command? Post 302078844 by vish_indian on Wednesday 5th of July 2006 03:53:24 AM
Old 07-05-2006
You may want to try something like this. This code prints the filename from the directory specified on command line.


Quote:
#! /bin/sh

for i in $@; do
echo "`awk -f awkvar2 $i`"
done
You can replace echo command with your function. awkvar2 file is:

NR==1{
print FILENAME;
}
"NR==1" is required to make it print the filename just once. BEGIN can't be used for the same because as per man page FILENAME variable isn't assigned a value till BEGIN block execution. What you need can be done without awk. This is just to display how to return variables to shell from awk.

Hope this helps.........
 

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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)
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