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Top Forums Shell Programming and Scripting Issue with AWK using a variable Post 302609833 by junnujayz on Tuesday 20th of March 2012 01:17:58 PM
Old 03-20-2012
Issue with AWK using a variable

Hi,

I am doing an AWK in ksh as below with the string to search to be read from variable but for some reason there is no output. It works when I hard code it.
Code:
awk 'substr($0,22,6)=="${VAR}"' XXX.txt' >YYY.txt

On reading other posts I tried below option,
Code:
'substr($0,22,6)=="/"${VAR}/""'

Also tried concatenating the quotes within the variable for ex if VAR=2011 I made it to VAR="2011".

When I did an echo of the awk the comment appears but the output is not generated. Please help.

Last edited by Franklin52; 03-20-2012 at 03:14 PM.. Reason: Please use code tags for data and code samples, thank you
 

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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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