Bash loop dynamic variable from source file - bash

Question

I got a config file called user.conf that contains this list:
USER1_USERNAME="John"
#USER2_USERNAME="Mike"
USER3_USERNAME="David"
USER4_USERNAME="James"
USER5_USERNAME="Jenny"
Notice that user can comment out that line USER2_USERNAME so, only 4 users are used in source file in bash script test.sh here:
#!/bin/bash
#source the username
source "user.conf"
n=1
while :; do
((n++))
if [ -n "${USER${n}_USERNAME}"]; then
echo "This variable USER${n}_USERNAME is set: ${USER${n}_USERNAME}"
else
echo "This variable USER${n}_USERNAME is not set"
fi
done
I want to display which variable is set with its value and skip the commented variables but at this point my code just display an error:
./test.sh: line 8: ${USER${n}_USERNAME}: bad substitution
Is it possible to loop that variable like above?
Also, user can define much more USER in that user.conf, for example
USER6_USERNAME="George "
Expected output:
This variable USER1_USERNAME is set to John
This variable USER2_USERNAME is not set
This variable USER3_USERNAME is set to David
This variable USER4_USERNAME is set to James
This variable USER5_USERNAME is set to Jenny

Answer 1

You can use bash variable substitution like this:
#!/usr/bin/env bash
source "user.conf"
for i in "${!USER#}"
do
echo $i ${!i}
done
Like your code, this first sources the users file, as it looks like a bash script. This of course has the potentially dangerous side effect that any other code in user.conf runs as well, so be careful and don't let strangers modify that file.
Then it uses ${!var#}, which expands to the names of variables whose names begin with a prefix, here "var", or for you "USER". You could also use ${!var*}, depending on whether you want all values in one quoted variable or multiple ones. See shell parameter expansion for details.
The whole approach is tied to a common prefix for your config variables. In this case, you'll also see $USER in the output, which is the name of the currently logged in user. You can filter that with e.g., grep or a simple if [ "$i" != "USER" ] in the loop.
If you want undefined variables as well, sourcing the users file may not be a good solution. You could instead read the file line by line and check for a leading #:
#!/usr/bin/env bash
set -eu
while IFS= read -r line
do
var=$(echo "$line" | cut -d '=' -f 1)
name=$(echo "$line" | cut -d '=' -f 2)
if [[ "$var" =~ ^# ]]
then
var=$(echo "$var" | cut -c 2-)
echo "The variable $var is not set"
else
echo "The variable $var is set to $name"
fi
done
Output:
bash users.sh < users.conf
The variable USER1_USERNAME is set to "John"
The variable USER2_USERNAME is not set
The variable USER3_USERNAME is set to "David"
The variable USER4_USERNAME is set to "James"
The variable USER5_USERNAME is set to "Jenny"
However, this approach is brittle as it doesn't understand bash syntax. A leading space would be fine when sourcing, but would trip the comment detection on this code. Variables in user names would not get expanded, which may or may not be a good thing.

Answer 2

I liked what Robert suggested. But IMHO whole concept of your script is wrong. This should be done with array(s) not dynamic vars.
users=(
"John"
"Mike"
"David"
"James"
"Jenny"
)
And than you just loop over that array
for user in "${users[#]}"; { echo "$user"; }
If you want to make this config editable by users this can also be done, thay can comment, delete or add items in array:
users=(
"John"
# "David"
"James"
"Jenny"
"Lenny"
)

Answer 3

I also found my own way to deal with this by modifying the source file user.conf to have suffix at the end:
#user.conf
USER_USERNAME1="John"
#USER_USERNAME2="Mike"
USER_USERNAME3="David"
USER_USERNAME4="James"
USER_USERNAME5="Jenny"
#!/bin/bash
source "user.conf"
usernames="${!USER_USERNAME#}"
username_count=$(echo "${usernames}" | wc -w)
count=1
while [[ ${count} -le ${username_count} ]]; do
typeset -n "username"="USER_USERNAME${count}"
if [ -n "${username}" ]; then
echo "The variable USER_USERNAME${count} is set to ${username}"
else
echo "The variable USER_USERNAME${count} is not set"
fi
((count = count + 1))
done
Since it's restricted to put this in question, I post it here.