Variable not expanding inside another variable bash

Question

I am reading a csv file in bash script as follows:

resource='/data/bscs/'
while IFS='|'read seqid fname fpath 
do
    echo "FILENO:         $seqid"
    echo "FILENAME:       $fname"
    echo "FILE_PATH:    $fpath"                    
done < "report.csv"

My csv file has following values :

3|sample1.txt|$resource/operation/

I want the $resource to expand inside $fname. Instead, I am getting this output:

FILENO:         3
FILENAME:       sample1.txt
SOURCE PATH:    $resource/operation

I have tried the following:

"${fpath}" $(echo $fpath)

How can I achieve this?

Answer 1

You can use eval, but the next maintainer will curse your name. Have you tried just appending the variable when using the parsed data, or expanding the variable when creating the data?

If the file is created that way, can't you just use some [redacted] techniques to convince the originator to change their wicked ways?

If change is literally not possible, then you must have control over which variables are possible. Otherwise your script is vulnerable to all sorts of injection attacks, such as inputs like 3|sample1.txt|$(rm --arr --eff /)/operation/. Since you obviously have that under control, you can do some literal replacements of variables with their values on a case by case basis:

IFS='/' read -a pathnames <<< "$fpath"
for pathname in "${pathnames[@]}"
do
    if [ "${pathname::1}" = '$' ]
    then
        variable_name="${pathname:1}"
        printf '%s' "${!variable_name}"
    else
        printf '%s' "$pathname"
    fi
done

With some additional boilerplate to add slashes between pathnames.

Answer 2

You can also use indirect variable expansion

$fpath=`echo $resource/operation`
echo ${!fpath}

This will expand the value of resource inside fpath.

Answer 3

Variable expansion happens just once and isn't recursive.

You can expand the value yourself:

resource=/data/bscs
fpath='$resource/operation'
echo ${fpath/'$resource'/$resource}

Note that this is much safer than eval, because all the possible variables to substitute must be listed in the code.

Answer 4

I do not know why people hate on eval. When used right does not cause problems.

Simplest solution would be:

    read -p "File Path: " FILE_PATH
    FILE_PATH=$(eval echo $FILE_PATH)

Edit: (explanation) Simply put, if you know your data will be OS PATH, just make sure it is expanded from variables before using it. Do that by evaluating all possible values in it, this is what eval is for. Using it before every command is just a nightmare - use it on your variables that you are using in command.