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How do I get the directory where a Bash script is located from within the script itself?
(74 answers)
Closed 6 years ago.
I have a Bash script that needs to know its full path. I'm trying to find a broadly-compatible way of doing that without ending up with relative or funky-looking paths. I only need to support Bash, not sh, csh, etc.
What I've found so far:
The accepted answer to Getting the source directory of a Bash script from within addresses getting the path of the script via dirname $0, which is fine, but that may return a relative path (like .), which is a problem if you want to change directories in the script and have the path still point to the script's directory. Still, dirname will be part of the puzzle.
The accepted answer to Bash script absolute path with OS X (OS X specific, but the answer works regardless) gives a function that will test to see if $0 looks relative and if so will pre-pend $PWD to it. But the result can still have relative bits in it (although overall it's absolute) — for instance, if the script is t in the directory /usr/bin and you're in /usr and you type bin/../bin/t to run it (yes, that's convoluted), you end up with /usr/bin/../bin as the script's directory path. Which works, but...
The readlink solution on this page, which looks like this:
# Absolute path to this script. /home/user/bin/foo.sh
SCRIPT=$(readlink -f $0)
# Absolute path this script is in. /home/user/bin
SCRIPTPATH=`dirname $SCRIPT`
But readlink isn't POSIX and apparently the solution relies on GNU's readlink where BSD's won't work for some reason (I don't have access to a BSD-like system to check).
So, various ways of doing it, but they all have their caveats.
What would be a better way? Where "better" means:
Gives me the absolute path.
Takes out funky bits even when invoked in a convoluted way (see comment on #2 above). (E.g., at least moderately canonicalizes the path.)
Relies only on Bash-isms or things that are almost certain to be on most popular flavors of *nix systems (GNU/Linux, BSD and BSD-like systems like OS X, etc.).
Avoids calling external programs if possible (e.g., prefers Bash built-ins).
(Updated, thanks for the heads up, wich) It doesn't have to resolve symlinks (in fact, I'd kind of prefer it left them alone, but that's not a requirement).
Here's what I've come up with (edit: plus some tweaks provided by sfstewman, levigroker, Kyle Strand, and Rob Kennedy), that seems to mostly fit my "better" criteria:
SCRIPTPATH="$( cd -- "$(dirname "$0")" >/dev/null 2>&1 ; pwd -P )"
That SCRIPTPATH line seems particularly roundabout, but we need it rather than SCRIPTPATH=`pwd` in order to properly handle spaces and symlinks.
The inclusion of output redirection (>/dev/null 2>&1) handles the rare(?) case where cd might produce output that would interfere with the surrounding $( ... ) capture. (Such as cd being overridden to also ls a directory after switching to it.)
Note also that esoteric situations, such as executing a script that isn't coming from a file in an accessible file system at all (which is perfectly possible), is not catered to there (or in any of the other answers I've seen).
The -- after cd and before "$0" are in case the directory starts with a -.
I'm surprised that the realpath command hasn't been mentioned here. My understanding is that it is widely portable / ported.
Your initial solution becomes:
SCRIPT=$(realpath "$0")
SCRIPTPATH=$(dirname "$SCRIPT")
And to leave symbolic links unresolved per your preference:
SCRIPT=$(realpath -s "$0")
SCRIPTPATH=$(dirname "$SCRIPT")
The simplest way that I have found to get a full canonical path in Bash is to use cd and pwd:
ABSOLUTE_PATH="$(cd "$(dirname "${BASH_SOURCE[0]}")" && pwd)/$(basename "${BASH_SOURCE[0]}")"
Using ${BASH_SOURCE[0]} instead of $0 produces the same behavior regardless of whether the script is invoked as <name> or source <name>.
I just had to revisit this issue today and found Get the source directory of a Bash script from within the script itself:
DIR="$( cd "$( dirname "${BASH_SOURCE[0]}" )" && pwd )"
There's more variants at the linked answer, e.g. for the case where the script itself is a symlink.
Get the absolute path of a shell script
It does not use the -f option in readlink, and it should therefore work on BSD/Mac OS X.
Supports
source ./script (When called by the . dot operator)
Absolute path /path/to/script
Relative path like ./script
/path/dir1/../dir2/dir3/../script
When called from symlink
When symlink is nested eg) foo->dir1/dir2/bar bar->./../doe doe->script
When caller changes the scripts name
I am looking for corner cases where this code does not work. Please let me know.
Code
pushd . > /dev/null
SCRIPT_PATH="${BASH_SOURCE[0]}";
while([ -h "${SCRIPT_PATH}" ]); do
cd "`dirname "${SCRIPT_PATH}"`"
SCRIPT_PATH="$(readlink "`basename "${SCRIPT_PATH}"`")";
done
cd "`dirname "${SCRIPT_PATH}"`" > /dev/null
SCRIPT_PATH="`pwd`";
popd > /dev/null
echo "srcipt=[${SCRIPT_PATH}]"
echo "pwd =[`pwd`]"
Known issus
The script must be on disk somewhere. Let it be over a network. If you try to run this script from a PIPE it will not work
wget -o /dev/null -O - http://host.domain/dir/script.sh |bash
Technically speaking, it is undefined. Practically speaking, there is no sane way to detect this. (A co-process can not access the environment of the parent.)
Use:
SCRIPT_PATH=$(dirname `which $0`)
which prints to standard output the full path of the executable that would have been executed when the passed argument had been entered at the shell prompt (which is what $0 contains)
dirname strips the non-directory suffix from a file name.
Hence you end up with the full path of the script, no matter if the path was specified or not.
As realpath is not installed per default on my Linux system, the following works for me:
SCRIPT="$(readlink --canonicalize-existing "$0")"
SCRIPTPATH="$(dirname "$SCRIPT")"
$SCRIPT will contain the real file path to the script and $SCRIPTPATH the real path of the directory containing the script.
Before using this read the comments of this answer.
Easy to read? Below is an alternative. It ignores symlinks
#!/bin/bash
currentDir=$(
cd $(dirname "$0")
pwd
)
echo -n "current "
pwd
echo script $currentDir
Since I posted the above answer a couple years ago, I've evolved my practice to using this linux specific paradigm, which properly handles symlinks:
ORIGIN=$(dirname $(readlink -f $0))
Simply:
BASEDIR=$(readlink -f $0 | xargs dirname)
Fancy operators are not needed.
You may try to define the following variable:
CWD="$(cd -P -- "$(dirname -- "${BASH_SOURCE[0]}")" && pwd -P)"
Or you can try the following function in Bash:
realpath () {
[[ $1 = /* ]] && echo "$1" || echo "$PWD/${1#./}"
}
This function takes one argument. If the argument already has an absolute path, print it as it is, otherwise print $PWD variable + filename argument (without ./ prefix).
Related:
Bash script absolute path with OS X
Get the source directory of a Bash script from within the script itself
Answering this question very late, but I use:
SCRIPT=$( readlink -m $( type -p ${0} )) # Full path to script handling Symlinks
BASE_DIR=`dirname "${SCRIPT}"` # Directory script is run in
NAME=`basename "${SCRIPT}"` # Actual name of script even if linked
We have placed our own product realpath-lib on GitHub for free and unencumbered community use.
Shameless plug but with this Bash library you can:
get_realpath <absolute|relative|symlink|local file>
This function is the core of the library:
function get_realpath() {
if [[ -f "$1" ]]
then
# file *must* exist
if cd "$(echo "${1%/*}")" &>/dev/null
then
# file *may* not be local
# exception is ./file.ext
# try 'cd .; cd -;' *works!*
local tmppwd="$PWD"
cd - &>/dev/null
else
# file *must* be local
local tmppwd="$PWD"
fi
else
# file *cannot* exist
return 1 # failure
fi
# reassemble realpath
echo "$tmppwd"/"${1##*/}"
return 0 # success
}
It doesn't require any external dependencies, just Bash 4+. Also contains functions to get_dirname, get_filename, get_stemname and validate_path validate_realpath. It's free, clean, simple and well documented, so it can be used for learning purposes too, and no doubt can be improved. Try it across platforms.
Update: After some review and testing we have replaced the above function with something that achieves the same result (without using dirname, only pure Bash) but with better efficiency:
function get_realpath() {
[[ ! -f "$1" ]] && return 1 # failure : file does not exist.
[[ -n "$no_symlinks" ]] && local pwdp='pwd -P' || local pwdp='pwd' # do symlinks.
echo "$( cd "$( echo "${1%/*}" )" 2>/dev/null; $pwdp )"/"${1##*/}" # echo result.
return 0 # success
}
This also includes an environment setting no_symlinks that provides the ability to resolve symlinks to the physical system. By default it keeps symlinks intact.
Considering this issue again: there is a very popular solution that is referenced within this thread that has its origin here:
DIR="$( cd "$( dirname "${BASH_SOURCE[0]}" )" && pwd )"
I have stayed away from this solution because of the use of dirname - it can present cross-platform difficulties, particularly if a script needs to be locked down for security reasons. But as a pure Bash alternative, how about using:
DIR="$( cd "$( echo "${BASH_SOURCE[0]%/*}" )" && pwd )"
Would this be an option?
If we use Bash I believe this is the most convenient way as it doesn't require calls to any external commands:
THIS_PATH="${BASH_SOURCE[0]}";
THIS_DIR=$(dirname $THIS_PATH)
The accepted solution has the inconvenient (for me) to not be "source-able":
if you call it from a "source ../../yourScript", $0 would be "bash"!
The following function (for bash >= 3.0) gives me the right path, however the script might be called (directly or through source, with an absolute or a relative path):
(by "right path", I mean the full absolute path of the script being called, even when called from another path, directly or with "source")
#!/bin/bash
echo $0 executed
function bashscriptpath() {
local _sp=$1
local ascript="$0"
local asp="$(dirname $0)"
#echo "b1 asp '$asp', b1 ascript '$ascript'"
if [[ "$asp" == "." && "$ascript" != "bash" && "$ascript" != "./.bashrc" ]] ; then asp="${BASH_SOURCE[0]%/*}"
elif [[ "$asp" == "." && "$ascript" == "./.bashrc" ]] ; then asp=$(pwd)
else
if [[ "$ascript" == "bash" ]] ; then
ascript=${BASH_SOURCE[0]}
asp="$(dirname $ascript)"
fi
#echo "b2 asp '$asp', b2 ascript '$ascript'"
if [[ "${ascript#/}" != "$ascript" ]]; then asp=$asp ;
elif [[ "${ascript#../}" != "$ascript" ]]; then
asp=$(pwd)
while [[ "${ascript#../}" != "$ascript" ]]; do
asp=${asp%/*}
ascript=${ascript#../}
done
elif [[ "${ascript#*/}" != "$ascript" ]]; then
if [[ "$asp" == "." ]] ; then asp=$(pwd) ; else asp="$(pwd)/${asp}"; fi
fi
fi
eval $_sp="'$asp'"
}
bashscriptpath H
export H=${H}
The key is to detect the "source" case and to use ${BASH_SOURCE[0]} to get back the actual script.
One liner
`dirname $(realpath $0)`
Bourne shell (sh) compliant way:
SCRIPT_HOME=`dirname $0 | while read a; do cd $a && pwd && break; done`
Perhaps the accepted answer to the following question may be of help.
How can I get the behavior of GNU's readlink -f on a Mac?
Given that you just want to canonicalize the name you get from concatenating $PWD and $0 (assuming that $0 is not absolute to begin with), just use a series of regex replacements along the line of abs_dir=${abs_dir//\/.\//\/} and such.
Yes, I know it looks horrible, but it'll work and is pure Bash.
Try this:
cd $(dirname $([ -L $0 ] && readlink -f $0 || echo $0))
I have used the following approach successfully for a while (not on OS X though), and it only uses a shell built-in and handles the 'source foobar.sh' case as far as I have seen.
One issue with the (hastily put together) example code below is that the function uses $PWD which may or may not be correct at the time of the function call. So that needs to be handled.
#!/bin/bash
function canonical_path() {
# Handle relative vs absolute path
[ ${1:0:1} == '/' ] && x=$1 || x=$PWD/$1
# Change to dirname of x
cd ${x%/*}
# Combine new pwd with basename of x
echo $(pwd -P)/${x##*/}
cd $OLDPWD
}
echo $(canonical_path "${BASH_SOURCE[0]}")
type [
type cd
type echo
type pwd
Just for the hell of it I've done a bit of hacking on a script that does things purely textually, purely in Bash. I hope I caught all the edge cases.
Note that the ${var//pat/repl} that I mentioned in the other answer doesn't work since you can't make it replace only the shortest possible match, which is a problem for replacing /foo/../ as e.g. /*/../ will take everything before it, not just a single entry. And since these patterns aren't really regexes I don't see how that can be made to work. So here's the nicely convoluted solution I came up with, enjoy. ;)
By the way, let me know if you find any unhandled edge cases.
#!/bin/bash
canonicalize_path() {
local path="$1"
OIFS="$IFS"
IFS=$'/'
read -a parts < <(echo "$path")
IFS="$OIFS"
local i=${#parts[#]}
local j=0
local back=0
local -a rev_canon
while (($i > 0)); do
((i--))
case "${parts[$i]}" in
""|.) ;;
..) ((back++));;
*) if (($back > 0)); then
((back--))
else
rev_canon[j]="${parts[$i]}"
((j++))
fi;;
esac
done
while (($j > 0)); do
((j--))
echo -n "/${rev_canon[$j]}"
done
echo
}
canonicalize_path "/.././..////../foo/./bar//foo/bar/.././bar/../foo/bar/./../..//../foo///bar/"
Yet another way to do this:
shopt -s extglob
selfpath=$0
selfdir=${selfpath%%+([!/])}
while [[ -L "$selfpath" ]];do
selfpath=$(readlink "$selfpath")
if [[ ! "$selfpath" =~ ^/ ]];then
selfpath=${selfdir}${selfpath}
fi
selfdir=${selfpath%%+([!/])}
done
echo $selfpath $selfdir
More simply, this is what works for me:
MY_DIR=`dirname $0`
source $MY_DIR/_inc_db.sh
I have a list of files that I can obtain using the UNIX 'find' command such as:
$ find . -name "*.txt"
foo/foo.txt
bar/bar.txt
How can I pass this output into a Python script like hello.py so I can parse it using Python's argparse library?
Thanks!
If you want just text output of find(1), then use a pipe:
~$ find . -name "*.txt" | python hello.py
If you are looking to pass list of files as arguments to the script, use xargs(1):
~$ find . -name "*.txt" -print0 | xargs -0 python hello.py
or use -exec option of find(1).
Use xargs:
find . -name "*.txt" | xargs python -c 'import sys; print sys.argv[1:]'
From man find:
-exec command ;
Execute command; true if 0 status is returned. All following
arguments to find are taken to be arguments to the command until
an argument consisting of `;' is encountered. The string `{}'
is replaced by the current file name being processed everywhere
it occurs in the arguments to the command, not just in arguments
where it is alone, as in some versions of find. Both of these
constructions might need to be escaped (with a `\') or quoted to
protect them from expansion by the shell. See the EXAMPLES sec‐
tion for examples of the use of the -exec option. The specified
command is run once for each matched file. The command is exe‐
cuted in the starting directory. There are unavoidable secu‐
rity problems surrounding use of the -exec action; you should
use the -execdir option instead.
-exec command {} +
This variant of the -exec action runs the specified command on
the selected files, but the command line is built by appending
each selected file name at the end; the total number of invoca‐
tions of the command will be much less than the number of
matched files. The command line is built in much the same way
that xargs builds its command lines. Only one instance of `{}'
is allowed within the command. The command is executed in the
starting directory.
So you can do
find . -name "*.txt" -exec python myscript.py {} +
This helps, if you need to pass arguments after the list of arguments from the find output:
$ python hello.py `find . -name "*.txt"`
I used it to concat pdf files into another one:
$ pdfunite `find . -name "*.pdf" | sort` all.pdf
I have a Django project and I'm working on Pylinting my way through it.
I have a couple situations where I'd like to be able to recursively find all files with a given name and pylint them differently (using different options). For example, I'd like to set different options for pylinting urls.py and admin.py
The following works for 1 directory..
pylint ./project_name/*/urls.py
But I'd like to make that * recursive... so that it drills into sub directories.
Any way to achieve that?
Update
I'd also like them all to run as a single pylint output, not sequentially
Depending on your operating system, you could use:
find project_name -name urls.py | xargs pylint
Try with find:
find ./project_name/ -name "urls.py" -exec pylint '{}' \;
If you want to run multiple files in a single call to pylint:
find ./project_name/ -name "urls.py" -exec pylint '{}' +
Get all python files (recursive)
Pass them all at once to pylint (from answer by Robin)
Show output in console and put it in a file
find . -name "*.py" -print0 | xargs -0 pylint 2>&1 | tee err_pylint.rst~