CODEWITHSUNDEEP
bash
a06e
a06e

Reputation: 20774

Get parent directory name and filename without extension from full filepath in bash?

I have a long file path, like

 /path/to/file/dir1/dir2/file.txt.gz

I am interested in getting the file name without the last extension (i.e., file.txt), and the name of the parent directory (dir2, I don't want the full path to the parent directory, just its name).

How can I do this in bash?

Upvotes: 1

Views: 697

Answers (2)

anubhava
anubhava

Reputation: 785276

Using BASH:

s='/path/to/file/dir1/dir2/file.txt.gz'

file="${s##*/}"
file="${file%.*}"
echo "$file"
file.txt

filder="${s%/*}"
folder="${folder##*/}"
echo "$folder"
dir2

Using awk:

awk -F '/' '{sub(/\.[^.]+$/, "", $NF); print $(NF-1), $NF}' <<< "$s"
dir2 file.txt

To read them into shell variables:

read folder file < <(awk -F '/' '{sub(/\.[^.]+$/, "", $NF);print $(NF-1), $NF}'<<<"$s")

Upvotes: 3

asjo
asjo

Reputation: 3194

The first part can be solved by basename(1):

$ basename /path/to/file/dir1/dir2/file.txt.gz
file.txt.gz
$

dirname(1) does the opposite, which is not quite what you want, but maybe you can use that as a starting point:

$ dirname /path/to/file/dir1/dir2/file.txt.gz
/path/to/file/dir1/dir2
$

Of course, you can always use Perl:

$ perl -E 'do { @p=split m|/|; say $p[-2] } for @ARGV' /path/to/file/dir1/dir2/file.txt.gz
dir2
$

Upvotes: 2

Related Questions