Check if Environment Variable exists for an user with su

Question

I'm on it since days, I need to check for an user $1 if an Environnement Variable $2 exists.

I tried some cases :

 $ [[ -v a ]] && echo "a ok" || echo "a no"
 a no
 $ [[ -v $HOME ]] && echo "a ok" || echo "a no"
 a no 

(So nope (I tried on Bash > 4.2))

if [ su - ${1} -c "[ -z "${2}" ]" ]

Now, I'm trying this approach because I don't find what's wrong :/ !

su - root -c "echo "$HOME" | wc | cut -d' ' -f2"
su - root -c "echo "$HOME" | wc | sed 's/ //g'"

Answer 1

The syntax in your first examples is incorrect. If you write:

[[ -v $HOME ]]

Then you are actually executing the command:

[[ -v /home/yourusername ]]

Which of course is going to fail. You want:

[[ -v HOME ]] && echo ok || echo no

You can combine that with su to get:

su - root -c "[[ -v HOME ]]" && echo ok || echo no

Your last examples won't work because you can't nest quotes, and because shell variable expansion happens before command execution. That is, given:

su - root -c "echo "$HOME" | wc | cut -d' ' -f2"

The command you execute is actually going to be:

su - root -c 'echo /home/somesuser | ...'

If you want the value of $HOME from the perspective of root, you would need to prevent variable expansion in the local shell, for example by using single quotes:

su - root -c 'echo $HOME | ...'

It's not clear to me why you're passing the output to wc, because the word count of $HOME is almost always going to be 1. Rather than post-processing the output of wc with sed or cut, you can just use the -w option to get the word count:

su - root -c 'echo $HOME | wc -w'