Replace Environment Variable in argument to shell script

Question

I've got a shell script that gets called with arguments that contain variable names. I want the variable to be replaced with their values. Consider the example below: It solves my problem, but it uses eval. I want to avoid eval for security reasons.

#!/bin/bash
#
# example:
# $> replace.sh arg '$VAR'
# arg value

VAR=value
ARGS=$(eval echo $*)

echo $ARGS

That has been bothering me for a long time now, too!!

Schnodderbalken
– Schnodderbalken

2015-10-22 13:12:10 +00:00
Commented Oct 22, 2015 at 13:12 — Schnodderbalken
– Schnodderbalken, Commented Oct 22, 2015 at 13:12
seen this? - stackoverflow.com/questions/10820343/…

Dummy00001
– Dummy00001

2015-10-22 13:25:52 +00:00
Commented Oct 22, 2015 at 13:25 — Dummy00001
– Dummy00001, Commented Oct 22, 2015 at 13:25

Gary_W · Accepted Answer · 2015-10-22 14:32:42Z

2

You are looking for indirect variable expansion:

$ SPY1=SPY2
$ SPY2="Lance Link"
$ SPY=${!SPY1}
$ echo $SPY
Lance Link
$

answered Oct 22, 2015 at 14:32

Gary_W

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Comments

jayme · Accepted Answer · 2016-05-11 13:42:27Z

1

Works for me with something like:

#!/bin/bash
#
# example:
# $> replace.sh arg '$VAR'
# arg value

VAR=value
ARGS=""
for arg in "$@"
do
  arg="${arg/$/}" # remove preceding $ for indirect variable expansion to work
  ARGS="${ARGS} ${!arg}"
done

echo "$ARGS"

answered May 11, 2016 at 13:42

jayme

1,32612 silver badges26 bronze badges

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