Bash: Dynamically create environment variables

Question

We're delivering secrets into our containers via a path, e.g.: /mnt/secrets

...and each subsequent file in that directory is the 'secret' with the contents being the value: e.g. /mnt/secrets/somepassword contains 'superdooperpassword123'

This is happening because we're managing our secrets from a secret-store provider (encrypted, etc.). However, I need to convert those files+filecontents to environment variables to be used in other scripts.

Here is the script I attempted to use to accomplish that, secrets.sh:

FILES="/mnt/secrets/*"
for f in $FILES
do
  FILE=$(basename $f)
  echo "Creating environment variable for the following secret:  $FILE"
  declare -xg $(echo $FILE)=$(cat $f)
done

And while this runs without error, I don't see (via 'set' or get nothing with 'echo $var') when I try to ensure they're available. I've tried multiple arguments for 'declare' but I can't seem to expose the declared variables running in secrets.sh (from the files and their contents) back to bash.

I know I'm probably missing something simple. Any assistance would be appreciate to direct me how to use a shell script that reads files and their contents from the filesystem into dynamically created environment variables. TIA!~

Don't use $(echo $FILE) -- it's an elaborate and error-prone way to do $FILE, because the $( ) and echo mostly cancel each other out. — Gordon Davisson
– Gordon Davisson, Commented Jan 27, 2022 at 18:38
@GordonDavisson, so how would you handle if the filename (that will end up as the var name) has hyphens in it that need to be swapped for underscores? I've presently modified to: declare -xg $(echo $FILE | tr '-' '_')=$(cat $f) -- NOTE: what I've posted is working for the 20(ish) secrets we're loading into our containers... — thepip3r
– thepip3r, Commented Jan 27, 2022 at 19:30
nm... if i need to, i can move it up into the $FILE var creation line. — thepip3r
– thepip3r, Commented Jan 27, 2022 at 19:47
Using echo this way to feed a pipe is ok, but for a simple substitution like this I'd use ${FILE//-/_}="$(cat $f)" (but note that this type of substitution is a bashism, and won't work in some more basic shells like dash). — Gordon Davisson
– Gordon Davisson, Commented Jan 27, 2022 at 20:41

Diego Torres Milano · Accepted Answer · 2022-01-27 18:09:32Z

1

You have to source the file

source secrets.sh

or

. secrets.sh

otherwise you are setting the variables of the child process which are not the same as the parent's

answered Jan 27, 2022 at 18:09

Diego Torres Milano

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