Basically I need to run the script with paths related to the shell script file location, how can I change the current directory to the same directory as where the script file resides?
In Bash, you should get what you need like this:
#!/usr/bin/env bash
BASEDIR=$(dirname "$0")
echo "$BASEDIR"
The original post contains the solution (ignore the responses, they don't add anything useful). The interesting work is done by the mentioned unix command readlink
with option -f
. Works when the script is called by an absolute as well as by a relative path.
For bash, sh, ksh:
#!/bin/bash
# Absolute path to this script, e.g. /home/user/bin/foo.sh
SCRIPT=$(readlink -f "$0")
# Absolute path this script is in, thus /home/user/bin
SCRIPTPATH=$(dirname "$SCRIPT")
echo $SCRIPTPATH
For tcsh, csh:
#!/bin/tcsh
# Absolute path to this script, e.g. /home/user/bin/foo.csh
set SCRIPT=`readlink -f "$0"`
# Absolute path this script is in, thus /home/user/bin
set SCRIPTPATH=`dirname "$SCRIPT"`
echo $SCRIPTPATH
See also: https://stackoverflow.com/a/246128/59087
readlink
. That's why I recommended using pushd/popd (built-ins for bash).
-f
option to readlink
does something different on OS X (Lion) and possibly BSD. stackoverflow.com/questions/1055671/…
-f
is not supported on OS X at all (as of Lion); there you can either drop the -f
to make do with resolving at most one level of indirection, e.g. pushd "$(dirname "$(readlink "$BASH_SOURCE" || echo "$BASH_SOURCE")")"
, or you can roll your own recursive symlink-following script as demonstrated in the linked post.
sh /some/other/directory/script.sh)
, in this case .
would be your pwd, not /some/other/directory
An earlier comment on an answer said it, but it is easy to miss among all the other answers.
When using bash:
echo this file: "$BASH_SOURCE"
echo this dir: "$(dirname "$BASH_SOURCE")"
Bash Reference Manual, 5.2 Bash Variables
dirname "$BASH_SOURCE"
instead to handle spaces in $BASH_SOURCE.
"$(dirname "${BASH_SOURCE[0]}")"
Assuming you're using bash
#!/bin/bash
current_dir=$(pwd)
script_dir=$(dirname "$0")
echo $current_dir
echo $script_dir
This script should print the directory that you're in, and then the directory the script is in. For example, when calling it from /
with the script in /home/mez/
, it outputs
/
/home/mez
Remember, when assigning variables from the output of a command, wrap the command in $(
and )
- or you won't get the desired output.
$(pwd)
is a lot of overhead compared to $PWD
, which POSIX explicitly requires to be set on shell startup and on cd
.
The best answer for this question was answered here:
Getting the source directory of a Bash script from within
And it is:
DIR="$( cd "$( dirname "${BASH_SOURCE[0]}" )" && pwd )"
One-liner which will give you the full directory name of the script no matter where it is being called from.
To understand how it works you can execute the following script:
#!/bin/bash
SOURCE="${BASH_SOURCE[0]}"
while [ -h "$SOURCE" ]; do # resolve $SOURCE until the file is no longer a symlink
TARGET="$(readlink "$SOURCE")"
if [[ $TARGET == /* ]]; then
echo "SOURCE '$SOURCE' is an absolute symlink to '$TARGET'"
SOURCE="$TARGET"
else
DIR="$( dirname "$SOURCE" )"
echo "SOURCE '$SOURCE' is a relative symlink to '$TARGET' (relative to '$DIR')"
SOURCE="$DIR/$TARGET" # if $SOURCE was a relative symlink, we need to resolve it relative to the path where the symlink file was located
fi
done
echo "SOURCE is '$SOURCE'"
RDIR="$( dirname "$SOURCE" )"
DIR="$( cd -P "$( dirname "$SOURCE" )" && pwd )"
if [ "$DIR" != "$RDIR" ]; then
echo "DIR '$RDIR' resolves to '$DIR'"
fi
echo "DIR is '$DIR'"
If you're using bash....
#!/bin/bash
pushd $(dirname "${0}") > /dev/null
basedir=$(pwd -L)
# Use "pwd -P" for the path without links. man bash for more info.
popd > /dev/null
echo "${basedir}"
pushd
/popd
with cd $(dirname "${0}")
and cd -
to make it work on other shells, if they have a pwd -L
.
As theMarko suggests:
BASEDIR=$(dirname $0)
echo $BASEDIR
This works unless you execute the script from the same directory where the script resides, in which case you get a value of '.'
To get around that issue use:
current_dir=$(pwd)
script_dir=$(dirname $0)
if [ $script_dir = '.' ]
then
script_dir="$current_dir"
fi
You can now use the variable current_dir throughout your script to refer to the script directory. However this may still have the symlink issue.
If you want to get the actual script directory (irrespective of whether you are invoking the script using a symlink or directly), try:
BASEDIR=$(dirname $(realpath "$0"))
echo "$BASEDIR"
This works on both linux and macOS. I couldn't see anyone here mention about realpath
. Not sure whether there are any drawbacks in this approach.
on macOS, you need to install coreutils
to use realpath
. Eg: brew install coreutils
.
ash
and dash
(e.g. on Alpine Linux)
cd $(dirname $(readlink -f $0))
BASE_DIR="$(cd "$(dirname "$0")"; pwd)";
echo "BASE_DIR => $BASE_DIR"
Let's make it a POSIX oneliner:
a="/$0"; a=${a%/*}; a=${a#/}; a=${a:-.}; BASEDIR=$(cd "$a"; pwd)
Tested on many Bourne-compatible shells including the BSD ones.
As far as I know I am the author and I put it into public domain. For more info see: https://www.jasan.tk/posts/2017-05-11-posix_shell_dirname_replacement/
cd: too many arguments
if spaces in the path, and returns $PWD
. (an obvious fix, but just shows how many edge cases there actually are)
INTRODUCTION
This answer corrects the very broken but shockingly top voted answer of this thread (written by TheMarko):
#!/usr/bin/env bash
BASEDIR=$(dirname "$0")
echo "$BASEDIR"
WHY DOES USING dirname "$0" ON IT'S OWN NOT WORK?
dirname $0 will only work if user launches script in a very specific way. I was able to find several situations where this answer fails and crashes the script.
First of all, let's understand how this answer works. He's getting the script directory by doing
dirname "$0"
$0 represents the first part of the command calling the script (it's basically the inputted command without the arguments:
/some/path/./script argument1 argument2
$0="/some/path/./script"
dirname basically finds the last / in a string and truncates it there. So if you do:
dirname /usr/bin/sha256sum
you'll get: /usr/bin
This example works well because /usr/bin/sha256sum is a properly formatted path but
dirname "/some/path/./script"
wouldn't work well and would give you:
BASENAME="/some/path/." #which would crash your script if you try to use it as a path
Say you're in the same dir as your script and you launch it with this command
./script
$0 in this situation will be ./script and dirname $0 will give:
. #or BASEDIR=".", again this will crash your script
Using:
sh script
Without inputting the full path will also give a BASEDIR="."
Using relative directories:
../some/path/./script
Gives a dirname $0 of:
../some/path/.
If you're in the /some directory and you call the script in this manner (note the absence of / in the beginning, again a relative path):
path/./script.sh
You'll get this value for dirname $0:
path/.
and ./path/./script (another form of the relative path) gives:
./path/.
The only two situations where basedir $0 will work is if the user use sh or touch to launch a script because both will result in $0:
$0=/some/path/script
which will give you a path you can use with dirname.
THE SOLUTION
You'd have account for and detect every one of the above mentioned situations and apply a fix for it if it arises:
#!/bin/bash
#this script will only work in bash, make sure it's installed on your system.
#set to false to not see all the echos
debug=true
if [ "$debug" = true ]; then echo "\$0=$0";fi
#The line below detect script's parent directory. $0 is the part of the launch command that doesn't contain the arguments
BASEDIR=$(dirname "$0") #3 situations will cause dirname $0 to fail: #situation1: user launches script while in script dir ( $0=./script)
#situation2: different dir but ./ is used to launch script (ex. $0=/path_to/./script)
#situation3: different dir but relative path used to launch script
if [ "$debug" = true ]; then echo 'BASEDIR=$(dirname "$0") gives: '"$BASEDIR";fi
if [ "$BASEDIR" = "." ]; then BASEDIR="$(pwd)";fi # fix for situation1
_B2=${BASEDIR:$((${#BASEDIR}-2))}; B_=${BASEDIR::1}; B_2=${BASEDIR::2}; B_3=${BASEDIR::3} # <- bash only
if [ "$_B2" = "/." ]; then BASEDIR=${BASEDIR::$((${#BASEDIR}-1))};fi #fix for situation2 # <- bash only
if [ "$B_" != "/" ]; then #fix for situation3 #<- bash only
if [ "$B_2" = "./" ]; then
#covers ./relative_path/(./)script
if [ "$(pwd)" != "/" ]; then BASEDIR="$(pwd)/${BASEDIR:2}"; else BASEDIR="/${BASEDIR:2}";fi
else
#covers relative_path/(./)script and ../relative_path/(./)script, using ../relative_path fails if current path is a symbolic link
if [ "$(pwd)" != "/" ]; then BASEDIR="$(pwd)/$BASEDIR"; else BASEDIR="/$BASEDIR";fi
fi
fi
if [ "$debug" = true ]; then echo "fixed BASEDIR=$BASEDIR";fi
PATH
, then you'd just cd
to the folder and get pwd
. But the usual need to get the path I guess is to be able to call other scripts relative to the path of the current, in which I fail to see how any of those cases could cause any issues.
This one-liner tells where the shell script is, does not matter if you ran it or if you sourced it. Also, it resolves any symbolic links involved, if that is the case:
dir=$(dirname $(test -L "$BASH_SOURCE" && readlink -f "$BASH_SOURCE" || echo "$BASH_SOURCE"))
By the way, I suppose you are using /bin/bash.
So many answers, all plausible, each with pro's and con's & slightly differeing objectives (which should probably be stated for each). Here's another solution that meets a primary objective of both being clear and working across all systems, on all bash (no assumptions about bash versions, or readlink
or pwd
options), and reasonably does what you'd expect to happen (eg, resolving symlinks is an interesting problem, but isn't usually what you actually want), handle edge cases like spaces in paths, etc., ignores any errors and uses a sane default if there are any issues.
Each component is stored in a separate variable that you can use individually:
# script path, filename, directory
PROG_PATH=${BASH_SOURCE[0]} # this script's name
PROG_NAME=${PROG_PATH##*/} # basename of script (strip path)
PROG_DIR="$(cd "$(dirname "${PROG_PATH:-$PWD}")" 2>/dev/null 1>&2 && pwd)"
With tcsh, you can use the :h
variable modifier to retrieve the path.
One caveat is that if the script is executed as tcsh myscript.csh
, then you will only get the script name. A workaround is to validate the path as shown below.
#!/bin/tcsh
set SCRIPT_PATH = $0:h
if ( $SCRIPT_PATH == $0 ) then
set SCRIPT_PATH = "."
endif
$SCRIPT_PATH/compile.csh > $SCRIPT_PATH/results.txt
More information about variable modifiers can be found at https://learnxinyminutes.com/docs/tcsh/
Basic version:
dir=$(dirname $0)
If the script might be called via $PATH
, then:
dir=$(dirname $(which $0))
If the script might be called like this: bash script.sh
, then:
dir=$(dirname $(which $0 2>/dev/null || realpath ./$0))
If you feel insanely unsafe, then:
dir="$(dirname -- "$(which -- "$0" 2>/dev/null || realpath -- "./$0")")"
one of the most robust ways i've found is
#!/bin/sh
relative_dir=`perl -e 'use Cwd "realpath";$pwd = realpath(shift); $pwd =~ s/\/[^\/]*$//; print $pwd' $0`
cd $relative_dir
works with symlinks and has worked for many of my coworkers no matter their choice of shell type
That should do the trick:
echo `pwd`/`dirname $0`
It might look ugly depending on how it was invoked and the cwd but should get you where you need to go (or you can tweak the string if you care how it looks).
`pwd`/`dirname $0`
but may still fail on symlinks
Success story sharing
readlink
as well (see al's answer below)$BASH_SOURCE
in lieu of$0
, because$0
doesn't always contain the path of the script being invoked, such as when 'sourcing' a script.$BASH_SOURCE
is Bash-specific, the question is about shell script in general.CUR_PATH=$(pwd)
orpwd
do return the current directory (which does not have to be the scripts parent dir)!$BASH_SOURCE
, and it returns what I needed. My script is being called from another script, and$0
returns.
while$BASH_SOURCE
returns the right subdirectory (in my casescripts
).