Retrieve parent directory of script

Question

I'm working on an uninstaller script to delete the parent folder where the script is installed.

/usr/local/Myapplication/Uninstaller/uninstall.sh

So uninstall.sh has to do this:

rm- rf /usr/local/Myapplication

I can retrieve the folder where uninstall resides

SYMLINKS=$(readlink -f "$0")
UNINSTALL_PATH=$(dirname "$SYMLINKS")

But I'm still unsure of the pretty way to get the parent path. I thought of using sed to demove the "Uninstaller" part of this path, but is there an elegant way to get the path to Myapplication folder to delete it?

Thank you

in your script: cd "$(dirname "$0")" && cd .. && cd .. && [ -d Myapplication/Uninstaller ] && rm -rf Myapplication (I added a check that the directory we are about to delete contains a subdir "Uninstaller", but you could maybe add a better check, for example of a necessary file within Myapplication ?). If you don't know Myapplication, then : cd "$(dirname $0)" && cd .. && zepath="$(pwd)" && cd .. && [ -f "${zepath}/somefilesthathouldbehere" ] && rm -rf "${zepath}" — Olivier Dulac
– Olivier Dulac, Commented Nov 25, 2013 at 17:24
See the famous Can a Bash script tell what directory it's stored in? question, then cd .. from there. — Dan Dascalescu
– Dan Dascalescu, Commented Aug 15, 2015 at 11:38

hidefromkgb · Accepted Answer · 2017-07-16 11:33:00Z

30

How about using dirname twice?

APP_ROOT="$(dirname "$(dirname "$(readlink -fm "$0")")")"

The quoting desaster is only necessary to guard against whitespace in paths. Otherwise it would be more pleasing to the eye:

APP_ROOT=$(dirname $(dirname $(readlink -fm $0)))

edited Jul 16, 2017 at 11:33

hidefromkgb

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answered Nov 25, 2013 at 14:54

Boldewyn

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3 Comments

MangO_O Over a year ago

This works but I'm kinda iffy on using dirname twice to acheive what I want. More for learning purpose is there a way to get it in a single command line? sure APP_ROOT=$(dirname "``$(readlink -f "$0")``") would work but is there any other way?

Boldewyn Over a year ago

You don't need to be iffy about that. dirname is tailored to give you the part of the directory of a path. It is exactly the right tool to do so. I update the answer for the syntax.

VasiliNovikov Over a year ago

Note, this method is insecure for "ugly" names. A name like "-rm -rf" will break the desired behavior. Probably "." will, too. I don't know if that's the best way, but I've created a separate "correctness-focused" question here: stackoverflow.com/questions/40700119/…

mihai_mandis · Accepted Answer · 2013-11-25 15:33:22Z

8

Just get the parent of the parent directory:

my_app_path=$(dirname $(dirname $(readlink -f "$0")))

answered Nov 25, 2013 at 15:33

mihai_mandis

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Comments

Abdennour TOUMI · Accepted Answer · 2020-03-06 08:05:39Z

5

I put this answer as comment at 2018. But since I got a great feedback about the effectiveness of the solution, I will share it here as well :

# dir of script 
DIR="$( cd "$( dirname "${BASH_SOURCE[0]}" )" && pwd )";
# parent dir of that dir
PARENT_DIRECTORY="${DIR%/*}"

answered Mar 6, 2020 at 8:05

Abdennour TOUMI

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1 Comment

BarryPye Over a year ago

Not a fan of changing the directory just to get some information that readlink could provide.

Sergey Fedorov · Accepted Answer · 2013-11-25 15:49:16Z

4

If you need an absolute path, then you need cd. Otherwise you can just use $(dirname $0)/..

cd $(dirname $0)/..
path=$(pwd)
cd - # go back

answered Nov 25, 2013 at 15:49

Sergey Fedorov

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Comments

thom · Accepted Answer · 2013-11-25 18:42:13Z

2

the ultimate simple way of getting the parent directory path:

PARENT_DIRECTORY="${PWD%/*}"

answered Nov 25, 2013 at 18:42

thom

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4 Comments

MangO_O Over a year ago

This is the parent directory of the current directory, not where the script resides. To use this I'd have to CD to the script directory first so it's not in a way too elegant.

Abdennour TOUMI Over a year ago

@MangO_O : So you need to replace PWD by script's dir : DIR="$( cd "$( dirname "${BASH_SOURCE[0]}" )" && pwd )"; PARENT_DIRECTORY="${DIR%/*}"

subelsky Over a year ago

@AbdennourTOUMI your solution worked well for me, seems easier than the accepted answer

Abdennour TOUMI Over a year ago

Welcome @subelsky. I've added it as answer to make it more visible for others: stackoverflow.com/a/60559856/747579

Joe Goggins · Accepted Answer · 2015-07-29 23:18:41Z

2

Full path to parent dir of script, i.e. "/usr/local/bin/bla": export PARENT_OF_THIS_SCRIPT=$( cd $(dirname $0) ; pwd -P )

Just the most recent parent of script, i.e. "bla": export PARENT_DIR_OF_SCRIPT=$( cd $(dirname $0) ; pwd -P | xargs basename )

answered Jul 29, 2015 at 23:18

Joe Goggins

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Comments

Pierre Arlaud · Accepted Answer · 2013-11-25 14:53:55Z

1

Why don't you simply add ../ at the end of the path?

answered Nov 25, 2013 at 14:53

Pierre Arlaud

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2 Comments

MangO_O Over a year ago

This wouldn't work with rm, UNINSTALL_PATH=$(dirname "$SYMLINKS")/../ is not the correct father path when used with the rm command

Boldewyn Over a year ago

Also you'd need to add /../. And if, for whatever reason, $UNINSTALL_PATH is empty, /../ == / and you're wiping the hard drive.

Abdull · Accepted Answer · 2022-01-04 18:53:14Z

1

As $0 can have suprising behavior, here is a solution using BASH_SOURCE[0]:

#/bin/bash

PARENT_DIR=$(dirname $(dirname $(readlink -f "${BASH_SOURCE[0]}")))

answered Jan 4, 2022 at 18:53

Abdull

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