Get the parent directory for a file

I want to get just the name of the parent directory for a file.

Example: When I have path=/a/b/c/d/file, I want only d and not /a/b/c/d (which I get from dirname $path) as output.

Is there any sophisticated way to do this?

6 Answers

It sounds like you want the basename of the dirname:

$ filepath=/a/b/c/d/file
$ parentname="$(basename "$(dirname "$filepath")")"
$ echo "$parentname"
d

4

You can use pwd to get the current working directory, and use parameter expansion to avoid forking it into another (sub)shell.

echo ${PWD##*/}

Edit: proven source

1

I think this is a less-resource solution:

 $ filepath=/a/b/c/d/file $ echo ${${filepath%/*}##*/} d

edit: Sorry, nested expansion isn't possible in bash, but it works in zsh. Bash-version:

 $ filepath=/a/b/c/d/file $ path=${filepath%/*} $ echo ${path##*/} d

2

In bash, in one line:

$ dirname /a/b/c/d/file | sed 's,^\(.*/\)\?\([^/]*\),\2,'

4

I like Julian67's answer above best, but here is a bit an expanded version:

file_path = "a/b/c/d/file.txt"
parent=$(echo $file_path | sed -e 's;\/[^/]*$;;') # cut away "/file.txt";'$' is end of string
parent=$(echo $parent | sed -e 's;.*\/;;') # cut away "/a/b/c/"
echo $parent # --> you get "d"

1

dirname and basename should be used for this task.

The short answer is

dirname /a/b/c/d/file | xargs basename
=> d

The first step, getting the dirname of the path, yields the path of the parent folder, as seen below.

dirname /a/b/c/d/file
=> /a/b/c/d

The path of the parent folder is then piped to another command with |. Since basename (and dirname) don't run on streams and instead just on parametized inputs, you cannot directly pipe the results to basename with | basename. This however is fixed by xargs, which turns the stream from the pipe into an input for basename

This allows the final solution to become

dirname /a/b/c/d/file | xargs basename

Which could also be written as ...

echo '/a/b/c/d/file' | xargs dirname | xargs basename

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