How to pass shell variable to awk command inside shell script

Question

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:

#!/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

Not sure I get it, $PFILE is the name of the file Awk should process on (or) just be passed to the body? — Inian
– Inian, Commented Sep 28, 2017 at 16:42
@Inian $PFILE is the name of file Awk should process as you said. Let's say this is inside the bash script named mybash.sh. Then I call the following command: bash mybash.sh -p=inputfile.txt — zwlayer
– zwlayer, Commented Sep 28, 2017 at 16:44
So why didn't awk '{print FILE $0}' ${PFILE} work? Remember FILE variable is undefined here. I am sure this would have printed the contents of inputfile.txt one line at a time — Inian
– Inian, Commented Sep 28, 2017 at 16:46
If you're calling it with an argument -p=inputfile.txt, then there's no need for the shift. You would shift if you called it with -p inputfile.txt, in which case the ${i#*=} doesn't work. Drop the shift. — William Pursell
– William Pursell, Commented Sep 28, 2017 at 17:01

randomir · Accepted Answer · 2017-10-04 13:57:36Z

2

To pass a shell variable to awk, you correctly used -v option.

However, the shift was unnecessary (you're iterating options with for), ;; was missing (you have to terminate each case branch), as well as was the name of the file for awk to process. Fixed, your script looks like:

#!/bin/bash
for i in "$@"; do
    case $i in
        -p=*|--producedfile=*)
            PFILE="${i#*=}"
            ;;
        *)
             # unknown option
            ;;
    esac
done

echo "PRODUCEDFILE = ${PFILE}"
awk  -v FILE="${PFILE}" '{print FILE, $0}' "${PFILE}"

Note however, awk already makes the name of the currently processed file available in the FILENAME variable. So, you could also write the last line as:

awk '{print FILENAME, $0}' "${PFILE}"

edited Oct 4, 2017 at 13:57

answered Sep 28, 2017 at 17:11

randomir

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