title: How to pass shell variable to awk command inside shell script
body: <p>I just want to pass a shell variable that stores  name of a file to awk command. When I searched this problem on the net I see many different options but none of them worked for me. I tried the followings:</p>
  
  <pre><code>#!/bin/bash
  for i in "$@"
  do
  case $i in
      -p=*|--producedfile=*)
      PFILE="${i#*=}"
      shift # past argument=value
      *)
            # unknown option
      ;;
  esac
  done
  echo "PRODUCEDFILE  = ${PFILE}"
  awk  -v FILE=${PFILE} '{print FILE $0}' #DIDNT WORK
  awk   '{print FILE $0}' ${PFILE} # DIDNT WORK
  awk  -v FILE=${PFILE} '{print $0}' FILE #DIDNT WORK
  </code></pre>