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Top Forums Shell Programming and Scripting Passing awk variable argument to a script which is being called inside awk Post 302769054 by RudiC on Monday 11th of February 2013 06:08:15 AM
Old 02-11-2013
Your idea is correct, but as your code is absolutely unreadable, I only can guess that the variable alDFid is undefined when used to build the external command.
And, of course, you have an unterminated string, i.e. one unmatched double quote (I guess in the printf statement)

Last edited by RudiC; 02-11-2013 at 07:10 AM.. Reason: typo / addendum
 

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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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