Bash bad substitution with subshell and substring

Question

A contrived example... given

FOO=\"/foo/bar/baz\"

this works (in bash)

BAR=$(basename $FOO) # result is BAR=\"baz\"
BAZ=${BAR         


        

Answer 1

Modified forms of parameter substitution such as ${parameter#word} can only modify a parameter, not an arbitrary word.

In this case, you might pipe the output of basename to a dd command, like

BAR=$(basename -- "$FOO" | dd bs=1 count=1 2>/dev/null)

(If you want a higher count, increase count and not bs, otherwise you may get fewer bytes than requested.)

In the general case, there is no way to do things like this in one assignment.