variable passed to awk Post: 55598

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variable passed to awk

Does anybody know how to print a variable passed to awk command? awk -F"|" 'BEGIN {print $job,"\n","Question \n"} {print $1,$2$4," ",$3}' "job=$job1" file1 I am trying to pass job the variable job1. the output is blank. ??

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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)

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