Extract directory from path

Question

In my script I need the directory of the file I am working with. For example, the file="stuff/backup/file.zip". I need a way to get the string "stuff/backup/" from the variable $file.

duplicates: [get parent directory of a file in bash](https://stackoverflow.com/q/40700119/995714), [Get the parent directory of a given file](https://unix.stackexchange.com/q/351916/44425), [Getting the parent of a directory in Bash](https://stackoverflow.com/q/8426058/995714) — phuclv, Mar 20 '21 at 01:21
@phuclv This question was asked on 2011-06-28, before all of those questions linked were posted (2016-11-20, 2017-03-16, and 2011-12-08, respectively). — Edwin, Jul 12 '21 at 17:21
@Edwin time is irrelevant on SO. [The question with a better set of answers remain open](https://meta.stackoverflow.com/q/251938/995714) — phuclv, Jul 13 '21 at 01:33

score 249 · Accepted Answer · answered Jun 28 '11 at 16:16

249

dirname $file

is what you are looking for

answered Jun 28 '11 at 16:16

Matthieu

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+1, beat me to it. would have been quicker but was prompoted to enter more than `dirname $file` – matchew Jun 28 '11 at 16:18
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`dirname "$file"`, rather. Without the quotes this will misbehave if the file or directory name has spaces. – Charles Duffy Sep 21 '18 at 16:38
1

fml, the question should be extract absolute dir from file, how do I do that? dirname yields relative path I guess – Alexander Mills Oct 19 '18 at 06:55
It is outstanding how difficult this is to accomplish in a win Batch script -- another reason not to use them I suppose! – bunkerdive Apr 22 '20 at 07:08

score 58 · Answer 2 · edited Jul 07 '16 at 19:30

58

dirname $file

will output

stuff/backup

which is the opposite of basename:

basename $file

would output

file.zip

edited Jul 07 '16 at 19:30

the Tin Man

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answered Jun 28 '11 at 16:18

matchew

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I figured Matt would be able to figure it from there (man dirname) :) – Matthieu Jun 28 '11 at 16:22
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I think we are all Matt's here. Just a guess. But I was just trying to distinguish my answer from yours. =) – matchew Jun 28 '11 at 17:29
Yeah, I do use basename a lot too, it's very nifty =) But I could just not find anything about returning the file's dir on google! I guess I wasn't looking up the right words. Hah, all 3 of us Matt's? :D – Matt Jun 28 '11 at 19:28
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Consider adding the necessary quotes to ensure that these commands work with names containing spaces (glob characters, etc). – Charles Duffy Sep 21 '18 at 16:39

score 7 · Answer 3 · answered Sep 21 '18 at 16:34

7

Using ${file%/*} like suggested by Urvin/LuFFy is technically better since you won't rely on an external command. To get the basename in the same way you could do ${file##*/}. It's unnecessary to use an external command unless you need to.

file="/stuff/backup/file.zip"
filename=${1##*/}     # file.zip
directory=${1%/*}     # /stuff/backup

It would also be fully POSIX compliant this way. Hope it helps! :-)

answered Sep 21 '18 at 16:34

MageParts

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There is *one* case where `dirname` has an advantage over the (more efficient) built-in approach, and that's if you aren't certain that your path is fully-qualified to start with. If you have `file=file.zip`, `dirname "$file"` will return `.`, whereas `${file%/*}` will return `file.zip`. – Charles Duffy Sep 21 '18 at 16:40
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...of course, you can branch: `case $file in */*) dir=${file%/*};; *) dir=.;; esac` is still POSIX-y and addresses the issue. – Charles Duffy Sep 21 '18 at 16:42

score 2 · Answer 4 · edited Jul 11 '18 at 05:46

2

For getting directorypath from the filepath:

file="stuff/backup/file.zip"
dirPath=${file%/*}/
echo ${dirPath}

edited Jul 11 '18 at 05:46

LuFFy

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answered Jul 11 '18 at 04:38

Urvin Shah

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score 1 · Answer 5 · answered Jun 28 '11 at 16:27

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Simply use $ dirname /home/~username/stuff/backup/file.zip

It will return /home/~username/stuff/backup/

answered Jun 28 '11 at 16:27

Nishchay Sharma

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Extract directory from path

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