How to extract directory path from file path?

Question

In Bash, if VAR="/home/me/mydir/file.c", how do I get "/home/me/mydir"?

A much more sophisticated and complex real directory path resolution is here https://stackoverflow.com/questions/29789204/bash-how-to-get-real-path-of-a-symlink/55254754#55254754 — Arunas Bartisius, Jul 23 '21 at 11:34

paxdiablo · Accepted Answer · 2021-08-28T23:57:15.507

917

dirname and basename are the tools you're looking for for extracting path components:

$ VAR='/home/pax/file.c'
$ DIR="$(dirname "${VAR}")" ; FILE="$(basename "${VAR}")"
$ echo "[${DIR}] [${FILE}]"
[/home/pax] [file.c]

They're not internal bash commands but they are part of the POSIX standard - see dirname and basename. Hence, they're probably available on, or can be obtained for, most platforms that are capable of running bash.

edited Aug 28 '21 at 23:57

answered May 25 '11 at 07:44

paxdiablo

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Why the use of brackets around the variable names, and not "$VAR" for example? – user658182 Jul 30 '17 at 12:34
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https://stackoverflow.com/questions/8748831/when-do-we-need-curly-braces-around-shell-variables answers the above question. – user658182 Jul 30 '17 at 12:34
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@user658182 In this particular example, it is done out of habit, not necessity. – Abandoned Cart Nov 14 '19 at 04:45
The `export` is unnecessary and the [`echo`s are useless.](http://www.iki.fi/era/unix/award.html#echo) – tripleee Aug 28 '21 at 18:09
2

@tripleee: the `export` is a habit of mine, simply to ensure the variable is passed to sub-shells. The `echo` statements are to show how you could get the output into a variable, but I should probably have gone the whole hog on that (which I now have). Though neither of those really affect the "meat" of the answer, I'll adjust. I'm always appreciative of constructive criticism on improving my answers. – paxdiablo Aug 28 '21 at 23:59

score 138 · Answer 2 · edited Feb 01 '17 at 23:07

138

$ export VAR=/home/me/mydir/file.c
$ export DIR=${VAR%/*}
$ echo "${DIR}"
/home/me/mydir

$ echo "${VAR##*/}"
file.c

To avoid dependency with basename and dirname

edited Feb 01 '17 at 23:07

Drew Noakes

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answered Feb 28 '13 at 14:19

Emmanuel Devaux

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Since both are part of POSIX a dependency should not be a problem. – orkoden Nov 15 '13 at 13:50
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orkoden , you're right. The aim of my answer is to show there is no obligation to execute two additional process. bash is self sufficient for the use case. – Emmanuel Devaux Nov 19 '13 at 08:59
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I am using Emmanuel's method because I wish to pass either a file or a folder name, and then compute the folder path. Using this regex does the right thing, whereas the dirname function returned the parent folder when I input a folder. – AnneTheAgile Dec 07 '13 at 19:33
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However, if there's no path info in $VAR, ${VAR%/*}/test produces an unexpected value equal to $VAR/test whereas $(dirname $VAR) will produce the more predictable and appropriate value of ./test. This is a big deal because the former will attempt to treat the filename as a directory while the latter will be OK. – davemyron Oct 01 '14 at 18:13
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The `export` is unnecessary here. The purpose of `export` is to make the variable visible in the environment of the shell's subprocesses, but you are not calling any subprocesses which use the environment to access this information. – tripleee Aug 28 '21 at 18:03
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This should arguably be the accepted answer. `dirname` and `basename` have their place, but if the path is already in a shell variable, using the shell's built-in facilities is more efficient and elegant than calling an external process. – tripleee Aug 28 '21 at 18:05
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Another benefit of this is it is much MUCH faster than `dirname` I'm running it in a loop with 1000 items and dirname takes seconds whereas this is nearly instantaneous – user1169420 Nov 02 '22 at 23:24

score 42 · Answer 3 · answered Jul 06 '16 at 09:13

42

On a related note, if you only have the filename or relative path, dirname on its own won't help. For me, the answer ended up being readlink.

fname='txtfile'    
echo $(dirname "$fname")                # output: .
echo $(readlink -f "$fname")            # output: /home/me/work/txtfile

You can then combine the two to get just the directory.

echo $(dirname $(readlink -f "$fname")) # output: /home/me/work

answered Jul 06 '16 at 09:13

jerblack

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if more than one path component not existed, you should use `readlink -m "$fname"` to canonicalize given name recursively – EDkan Sep 20 '16 at 12:13

score 7 · Answer 4 · answered Feb 26 '16 at 01:16

7

If you care target files to be symbolic link, firstly you can check it and get the original file. The if clause below may help you.

if [ -h $file ]
then
 base=$(dirname $(readlink $file))
else
 base=$(dirname $file)
fi

answered Feb 26 '16 at 01:16

Tahsin Turkoz

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aerijman · Answer 5 · 2020-05-23T17:05:21.657

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HERE=$(cd $(dirname $BASH_SOURCE) && pwd)

where you get the full path with new_path=$(dirname ${BASH_SOURCE[0]}). You change current directory with cd new_path and then run pwd to get the full path to the current directory.

edited May 23 '20 at 17:05

answered May 23 '20 at 16:57

aerijman

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Brilliant, the most polymorphic option! – Lucky Brain Nov 26 '20 at 03:02
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This seems to be an answer to a different question actually. [The quoting is broken.](https://stackoverflow.com/questions/10067266/when-to-wrap-quotes-around-a-shell-variable) – tripleee Aug 29 '21 at 06:28

score 4 · Answer 6 · answered Jun 07 '17 at 01:06

I was playing with this and came up with an alternative.

$ VAR=/home/me/mydir/file.c

$ DIR=`echo $VAR |xargs dirname`

$ echo $DIR
/home/me/mydir

The part I liked is it was easy to extend backup the tree:

$ DIR=`echo $VAR |xargs dirname |xargs dirname |xargs dirname`

$ echo $DIR
/home

alper · Answer 7 · 2021-07-25T16:05:19.957

2

You could try something like this using approach for How to find the last field using 'cut':

Explanation

rev reverses /home/user/mydir/file_name.c to be c.eman_elif/ridym/resu/emoh/
cut uses / as the delimiter, and chooses the second field, which is ridym/resu/emoh/, which deletes string up to the first occurrence of /
lastly, we reverse it again to get /home/user/mydir

$ VAR="/home/user/mydir/file_name.c"
$ echo $VAR | rev | cut -d"/" -f2- | rev
/home/user/mydir

edited Jul 25 '21 at 16:05

answered Mar 15 '19 at 21:12

alper

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Replacing '/' (not valid within a file name) with space (valid within a file name) is not recommended. If you must use such a solution, better to just remove everything up until, and including, the last '/' character. – Martyn Davis Dec 16 '20 at 00:10
@MartynDavis Please see my updated answer based on your suggestion – alper Dec 16 '20 at 21:22
This is not what the OP requested. – OldBuildingAndLoan Jul 25 '21 at 15:40
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@m42 updated my answer according to OP requested. Please see its updated version. – alper Jul 25 '21 at 16:05
@alper; yeah now that's what the OP requested. :) – OldBuildingAndLoan Jul 28 '21 at 17:05
@m42 Thanks for the heads up – alper Jul 28 '21 at 19:59

score 1 · Answer 8 · edited Aug 28 '21 at 18:06

1

Here is a script I used for recursive trimming. Replace $1 with the directory you want, of course.

BASEDIR=$1
IFS=$'\n'
cd "$BASEDIR"
 for f in $(find . -type f -name ' *')
 do 
    DIR=$(dirname "$f")
    DIR=${DIR:1}
    cd "$BASEDIR$DIR"
    rename 's/^ *//' *
 done

edited Aug 28 '21 at 18:06

tripleee

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answered Jun 18 '17 at 10:43

kmchen

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Using a `for` loop over the output of `find` is an antipattern and a source of many bugs. This construct is inherently limited, and will fail if `find` produces results which contain whitespace or other shell metacharacters, let alone then newlines. – tripleee Aug 28 '21 at 18:08

score 0 · Answer 9 · answered Aug 12 '22 at 10:38

0

I like my paths absolute, and I need it from the (first) bash parameter (not a variable, not $pdf, not script location):

i.e. to reliably create a subfolder next to my chosen file ($1):

rp="$(realpath $(dirname $1)/pfd_extract)"
mkdir -p "$rp"

answered Aug 12 '22 at 10:38

Frank N

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How to extract directory path from file path?

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