Im trying to write a script:
env PYTHONPATH=$PYTHONPATH: $Dir/scripts find * -name ‘*.py' -exec pylint (} \\; | grep . && exit 1
The code is finding all scripts in the root directory instead of using the environmental variables I set. Any help on writing this code to only look for files in the directory I set as a value in PYTHONPATH.
env PYTHONPATH=$PYTHONPATH: $Dir/scripts isn't doing what you think it's doing. Including $PYTHONPATH includes the former value of PYTHONPATH, meaning whatever you have it already set to or a blank default. The space in your variable also makes it invalid, and instead interprets the $Dir/scripts as a new command. It looks like what you want would be env PYTHONPATH=$Dir/scripts — but there's actually an easier way.
If you have __init__.py files in your directory, you can just do pylint ./some-directory. If you don't, you can use xargs: find . -type f -name "*.py" | xargs pylint. If you wanted to pass the directory instead of have it coded to . (your current calling directory) you could do that too:
# set directory to first argument
dir="$1"
# check if "dir" was actually provided, if not, set to `.`
if [ -z "$dir" ]; then dir=.; fi
find "$dir" -type f -name "*.py" | xargs pylint
You could save that in a script or function and then call it either with a directory (like run-pylint-on-everything.sh ~/foo/bar, or not, in which case it would run starting from your current shell location.
There’s no space between the PYTHONPATH value, it was a typo mistake, I want to run the command on a CLI instead of a script.
I'm trying to copy some files from one directory to another. I want all files in one directory to end up in the root of another directory.
This command does exactly what I want when I run it in the terminal:
cp -rv ./src/CopyPasteIntoBuildDir/* ./build-root/src/
This line of python, however, copies most of the files just like the above command, but it leaves some of the new files empty. Specifically, files in subdirectories are left empty.
subprocess.check_call("cp -rv ./src/CopyPasteIntoBuildDir/* ./build-root/src/", shell=True)
It creates the files if they're not there, and it truncates them if they are.
What is going on?
Assuming that you're decided to use cp rather than native Python operations --
This code will be much more reliable if you write it to not invoke any shell whatsoever. To avoid the need for /* on the source (and the side effects of this -- ie. refusal to copy directories whose names exceed the ARG_MAX combined environment and command-line size storage limit), use . as the last element of the name of the directory whose contents are to be copied, instead of passing a wildcard that needs to be expanded by a shell.
subprocess.check_call(["cp", "-R", "--", '%s/.' % src, dest])
The use of cp -R rather than cp -rv is on account of -R, but not -r, being POSIX-standardized (and thus portable across all compliant UNIXlike platforms).
Demonstrating In Action (copy/pasteable code)
tempdir=$(mktemp -d -t testdir.XXXXXX)
trap 'rm -rf "$tempdir"' EXIT
cd "$tempdir"
mkdir -p ./src/CopyPasteIntoBuildDir/subdir-1 ./build-root/src/
touch ./src/CopyPasteIntoBuildDir/file-1
touch ./src/CopyPasteIntoBuildDir/subdir-1/file-2
script='
import sys, shutil, subprocess
src = sys.argv[1]
dest = sys.argv[2]
subprocess.check_call(["cp", "-R", "--", "%s/." % src, dest])
'
python -c "$script" ./src/CopyPasteIntoBuildDir ./build-root/src/
find ./build-root -type f -print
rm -rf "$tempdir"
...emits output akin to:
./build-root/src/file-1
./build-root/src/subdir-1/file-2
...showing that content was correctly recursively copied with no prefix.
So apparently this is a problem with sh. Using bash instead worked.
subprocess.check_call("cp -rv ./src/CopyPasteIntoBuildDir/* ./build-root/src/", shell=True, executable="/bin/bash")
EDIT: See accepted answer!
This question already has answers here:
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 just moved to Linux (Ubuntu 15.04) and I am trying to add my python directory (:~/Documents/Python/Programs) to the path variable, but I am struggling..
I have tried export PATH = $PATH:~/Documents/Python/Programs, and then turning off and on again, but nothing happens
I have also looked at my ~/.profile, but don't under stand it, it comes up with (I have removed a tonnes of comments from the top):
if [ -n "$BASH_VERSION" ]; then
# include .bashrc if it exists
if [ -f "$HOME/.bashrc" ]; then
. "$HOME/.bashrc"
fi
fi
# set PATH so it includes user's private bin if it exists
if [ -d "$HOME/bin" ] ; then
PATH="$HOME/bin:$PATH"
fi
UPDATE:
What I am trying to do is add my python directory to PATH so that I will be able to import self made modules from within this directory
I was under the impression I had to add this to PATH by adding
PATH="$HOME/Documents/Python/Programs/:$PATH"
To the bottom of my ~/.profile document, was this wrong, and what should I actually be doing to solve this?
You have to set your PATH e.g. inside ~/.profile or ~/.bashrc depending on where you want to use Python. Add something like this to the end of either of them:
PATH="$HOME/Documents/Python/Programs/:$PATH"
As stated by the comments, this change will only be take into account, after either restarting a login shell or new start of X session (e.g. newboot). If you need the changes directly, either source the file or export it manually... so
either
source ~/.profile
source ~/.bashrc
(you could also use the . operator, but this is only working in Bash)
or export the variable
export PATH="$HOME/Documents/Python/Programs/:$PATH"
It's important to add your custom path before original PATH as the shell will call the 1st file found inside $PATH.