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Top Forums Shell Programming and Scripting Var substitution in awk - not working as expected Post 95653 by videsh77 on Friday 13th of January 2006 05:55:45 AM
Old 01-13-2006
Var substitution in awk - not working as expected

countA=`awk '/X/''{print substr($0,38,1)}' fName | wc -l`
countB=`wc -l fName | awk '{print int($1)}'`
echo > temp
ratio=`awk -va=$countA -vc=$countB '{printf "%.4f", a/c}' temp`

After running script for above I am getting an error as :

awk: 0602-533 Cannot find or open file -vc=25.
The source line number is 1.


Same kind of script is working elsewhere. Not sure why its going wrong.
 

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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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