Bash – Get current directory name (without full path) in a Bash script

bashshell

How would I get just the current working directory name in a bash script, or even better, just a terminal command.

pwd gives the full path of the current working directory, e.g. /opt/local/bin but I only want bin

Best Answer

No need for basename, and especially no need for a subshell running pwd (which adds an extra, and expensive, fork operation); the shell can do this internally using parameter expansion:

result=${PWD##*/}          # to assign to a variable

printf '%s\n' "${PWD##*/}" # to print to stdout
                           # ...more robust than echo for unusual names
                           #    (consider a directory named -e or -n)

printf '%q\n' "${PWD##*/}" # to print to stdout, quoted for use as shell input
                           # ...useful to make hidden characters readable.

Note that if you're applying this technique in other circumstances (not PWD, but some other variable holding a directory name), you might need to trim any trailing slashes. The below uses bash's extglob support to work even with multiple trailing slashes:

dirname=/path/to/somewhere//
shopt -s extglob           # enable +(...) glob syntax
result=${dirname%%+(/)}    # trim however many trailing slashes exist
result=${result##*/}       # remove everything before the last / that still remains
printf '%s\n' "$result"

Alternatively, without extglob:

dirname="/path/to/somewhere//"
result="${dirname%"${dirname##*[!/]}"}" # extglob-free multi-trailing-/ trim
result="${result##*/}"                  # remove everything before the last /
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