Bash: Variable variable name and default fallback

bash_question.md

I'm trying to write in bash what would look like this in PHP 7:

echo ${$branch . '_SSH'} ?? $default_SSH;

1. Check for a variable which has a variable name.
2. Fall back to another variable if it doesn't exist.
3. echo the result.

Is there an equivalent one liner in bash?

this might be helping, although i'm not sure if it's the shortest/best solution::

SSH=$([ $BRANCH ] && echo $BRANCH"_SSH" || echo 'default_SSH') && echo ${!SSH}

however, always avoid to write "clever" code. be explizit and simple. Future-you will appreciate. ;-)

Thanks for the input. This doesn't kill it 100%. It checks if $BRANCH exists but not if the variable ${$BRANCH'_SSH'} exists.