Exporting variables in a loop

Question

I need some help with my script.

I have environmental variables like these:

SECRET_OF_AN_APP_dev=123
SECRET_OF_AN_APP_test=456
SECRET_OF_AN_APP_prod=789

I want to have a script that will change SECRET_OF_AN_APP_dev=123 to SECRET_OF_AN_APP=123 and export it into env in first iteration, then SECRET_OF_AN_APP_test=456 change to SECRET_OF_AN_APP=456 in second iteration etc.

#!/bin/bash
APP_SECRET=SECRET_OF_AN_APP
ENVIRONMENT_NAME=dev
LIST_OF_ENVIRONMENTS=[dev, prod, test...]
for i in $LIST_OF_ENVIRONMENTS
do
APP_SECRET=$(env | grep APP_SECRET_$i | sed 's/APP_SECRET_'"$i"'=//g' ) >> var.bash
. ./var.bash
./some_script_where_$APP_SECRET_is_required.sh
done

I think my script is working in something like a subshell because when I do: echo $SECRET_OF_AN_APP, it's blank.

I need to have it set because other scripts which are executing in this loop need those variables.

I tried with export, with ./var.bash, with set -o allexport, etc.

Nothing is working.

I need to have "echo $SECRET_OF_AN_APP" in my shell, not only script. It's working in script. but it's not set in my environmental variables.

Answer 1

One piece that is tripping you up is that your env | grep APP_SECRET_$i is not substituting the value of APP_SECRET to the expected variable name of SECRET_OF_AN_APP. Add a variable expansion to APP_SECRET to fix that.

You've captured the output of $(env ...) into an assignment APP_SECRET and then redirected the output of that assignment to the file. The output from the command substitution may be correct, and assigned to APP_SECRET, but the text of that assignment is not present in the output file.

Either export APP_SECRET (since it's been assigned in this shell script), or print the assignment into the file before you source it and skip the redirection on the assignment.

Either:

export APP_SECRET

or:

APP_SECRET=$(varname=${APP_SECRET}_${i}; printf '%s' "${!varname}" )
declare -p APP_SECRET >> var.bash
echo "export APP_SECRET" >> var.bash # if you need it to be exported 

I've simplified your command substitution so that it puts together a variable that holds the name of the actual variable value that you want. Be careful to use braces here so that the underscore isn't seen as part of the APP_SECRET variable. Then we use bash's variable indirection expansion to expand "varname" into the name of the variable and then take the value of that variable. This translates "varname" into "SECRET_OF_AN_APP_dev", for example, and then looks for the value of that (e.g. 123).

If the value of APP_SECRET needs to be available to your shell after this script exits, then your parent shell needs to source in the var.bash file:

. ./var.bash

See also:

• How to refer variable defined in sub shell in parent shell
• How to make a variable from a subshell available in the parent shell
• Is it possible to pass environment variables from child to parent in user space?

Answer 2

Is this what are you looking for?

#!/bin/bash
SECRET_OF_AN_APP_dev=123
SECRET_OF_AN_APP_test=456
SECRET_OF_AN_APP_prod=789
for s in dev test prod
do
   name=SECRET_OF_AN_APP_$s
   echo $name
   SECRET_OF_AN_APP=${!name}
   echo $SECRET_OF_AN_APP
   export SECRET_OF_AN_APP
   some_app_which_uses_that_variable
done