Question

I can’t seem to “access” the value of my loop variable when executing a command line argument in a Bash script. I’d like to be able to write something like

#!/bin/bash
for myvar in 1 2 3
do
  $@
done

and run the script as ./myscript echo "${myvar}".

When I do this, the lines are echoed as empty. I probably don’t have a firm grasp one exactly what’s being evaluated where.

Is what I want even possible?

Answer 1

$myvar is evaluated before the child script is even run, so it can’t be evaluated within.

That is, when you invoke your script as:

./myscript echo "${myvar}"

what is actually being called is:

./myscript echo ''

presuming that $myvar is empty in the enclosing environment.

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If you wanted to be evil (and this is evil, and will create bugs, and you should not do it), you could use eval:

#!/bin/bash
for myvar in 1 2 3; do
  eval "$1"
done

…and then call as:

./myscript 'echo "${myvar}"'

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Something less awful would be to export a variable:

#!/bin/bash
export myvar
for myvar in 1 2 3; do
  "$@"
done

…but even then, the way you call the script will need to avoid preevaluation. For instance, this will work in conjunction with the above:

./myscript bash -c 'echo "$myvar"'

…but that basically has you back to eval. On the other hand:

./myscript ./otherscript

…will do what you want, presuming that otherscript refers to $myvar somewhere within. (This need not be a shell script, or even a script; it can be an executable or other command, and it will still be able to find myvar in the environment).

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What’s the real goal (ie. business purpose) you’re trying to achieve? There’s probably a way to accomplish it better aligned with best practices.