Equivalent of %~dp0 (retrieving source file name) in sh - linux

I'm converting some Windows batch files to Unix scripts using sh. I have problems because some behavior is dependent on the %~dp0 macro available in batch files.
Is there any sh equivalent to this? Any way to obtain the directory where the executing script lives?

The problem (for you) with $0 is that it is set to whatever command line was use to invoke the script, not the location of the script itself. This can make it difficult to get the full path of the directory containing the script which is what you get from %~dp0 in a Windows batch file.
For example, consider the following script, dollar.sh:
#!/bin/bash
echo $0
If you'd run it you'll get the following output:
# ./dollar.sh
./dollar.sh
# /tmp/dollar.sh
/tmp/dollar.sh
So to get the fully qualified directory name of a script I do the following:
cd `dirname $0`
SCRIPTDIR=`pwd`
cd -
This works as follows:
cd to the directory of the script, using either the relative or absolute path from the command line.
Gets the absolute path of this directory and stores it in SCRIPTDIR.
Goes back to the previous working directory using "cd -".

Yes, you can! It's in the arguments. :)
look at
${0}
combining that with
{$var%Pattern}
Remove from $var the shortest part of $Pattern that matches the back end of $var.
what you want is just
${0%/*}
I recommend the Advanced Bash Scripting Guide
(that is also where the above information is from).
Especiall the part on Converting DOS Batch Files to Shell Scripts
might be useful for you. :)
If I have misunderstood you, you may have to combine that with the output of "pwd". Since it only contains the path the script was called with!
Try the following script:
#!/bin/bash
called_path=${0%/*}
stripped=${called_path#[^/]*}
real_path=`pwd`$stripped
echo "called path: $called_path"
echo "stripped: $stripped"
echo "pwd: `pwd`"
echo "real path: $real_path
This needs some work though.
I recommend using Dave Webb's approach unless that is impossible.

In bash under linux you can get the full path to the command with:
readlink /proc/$$/fd/255
and to get the directory:
dir=$(dirname $(readlink /proc/$$/fd/255))
It's ugly, but I have yet to find another way.

I was trying to find the path for a script that was being sourced from another script. And that was my problem, when sourcing the text just gets copied into the calling script, so $0 always returns information about the calling script.
I found a workaround, that only works in bash, $BASH_SOURCE always has the info about the script in which it is referred to. Even if the script is sourced it is correctly resolved to the original (sourced) script.

The correct answer is this one:
How do I determine the location of my script? I want to read some config files from the same place.
It is important to realize that in the general case, this problem has no solution. Any approach you might have heard of, and any approach that will be detailed below, has flaws and will only work in specific cases. First and foremost, try to avoid the problem entirely by not depending on the location of your script!
Before we dive into solutions, let's clear up some misunderstandings. It is important to understand that:
Your script does not actually have a location! Wherever the bytes end up coming from, there is no "one canonical path" for it. Never.
$0 is NOT the answer to your problem. If you think it is, you can either stop reading and write more bugs, or you can accept this and read on.
...

Try this:
${0%/*}

This should work for bash shell:
dir=$(dirname $(readlink -m $BASH_SOURCE))
Test script:
#!/bin/bash
echo $(dirname $(readlink -m $BASH_SOURCE))
Run test:
$ ./somedir/test.sh
/tmp/somedir
$ source ./somedir/test.sh
/tmp/somedir
$ bash ./somedir/test.sh
/tmp/somedir
$ . ./somedir/test.sh
/tmp/somedir

This is a script can get the shell file real path when executed or sourced.
Tested in bash, zsh, ksh, dash.
BTW: you shall clean the verbose code by yourself.
#!/usr/bin/env bash
echo "---------------- GET SELF PATH ----------------"
echo "NOW \$(pwd) >>> $(pwd)"
ORIGINAL_PWD_GETSELFPATHVAR=$(pwd)
echo "NOW \$0 >>> $0"
echo "NOW \$_ >>> $_"
echo "NOW \${0##*/} >>> ${0##*/}"
if test -n "$BASH"; then
echo "RUNNING IN BASH..."
SH_FILE_RUN_PATH_GETSELFPATHVAR=${BASH_SOURCE[0]}
elif test -n "$ZSH_NAME"; then
echo "RUNNING IN ZSH..."
SH_FILE_RUN_PATH_GETSELFPATHVAR=${(%):-%x}
elif test -n "$KSH_VERSION"; then
echo "RUNNING IN KSH..."
SH_FILE_RUN_PATH_GETSELFPATHVAR=${.sh.file}
else
echo "RUNNING IN DASH OR OTHERS ELSE..."
SH_FILE_RUN_PATH_GETSELFPATHVAR=$(lsof -p $$ -Fn0 | tr -d '\0' | grep "${0##*/}" | tail -1 | sed 's/^[^\/]*//g')
fi
echo "EXECUTING FILE PATH: $SH_FILE_RUN_PATH_GETSELFPATHVAR"
cd "$(dirname "$SH_FILE_RUN_PATH_GETSELFPATHVAR")" || return 1
SH_FILE_RUN_BASENAME_GETSELFPATHVAR=$(basename "$SH_FILE_RUN_PATH_GETSELFPATHVAR")
# Iterate down a (possible) chain of symlinks as lsof of macOS doesn't have -f option.
while [ -L "$SH_FILE_RUN_BASENAME_GETSELFPATHVAR" ]; do
SH_FILE_REAL_PATH_GETSELFPATHVAR=$(readlink "$SH_FILE_RUN_BASENAME_GETSELFPATHVAR")
cd "$(dirname "$SH_FILE_REAL_PATH_GETSELFPATHVAR")" || return 1
SH_FILE_RUN_BASENAME_GETSELFPATHVAR=$(basename "$SH_FILE_REAL_PATH_GETSELFPATHVAR")
done
# Compute the canonicalized name by finding the physical path
# for the directory we're in and appending the target file.
SH_SELF_PATH_DIR_RESULT=$(pwd -P)
SH_FILE_REAL_PATH_GETSELFPATHVAR=$SH_SELF_PATH_DIR_RESULT/$SH_FILE_RUN_BASENAME_GETSELFPATHVAR
echo "EXECUTING REAL PATH: $SH_FILE_REAL_PATH_GETSELFPATHVAR"
echo "EXECUTING FILE DIR: $SH_SELF_PATH_DIR_RESULT"
cd "$ORIGINAL_PWD_GETSELFPATHVAR" || return 1
unset ORIGINAL_PWD_GETSELFPATHVAR
unset SH_FILE_RUN_PATH_GETSELFPATHVAR
unset SH_FILE_RUN_BASENAME_GETSELFPATHVAR
unset SH_FILE_REAL_PATH_GETSELFPATHVAR
echo "---------------- GET SELF PATH ----------------"
# USE $SH_SELF_PATH_DIR_RESULT BEBLOW

I have tried $0 before, namely:
dirname $0
and it just returns "." even when the script is being sourced by another script:
. ../somedir/somescript.sh

Related

How to create a command in linux from a bash executable when my program uses an internal database? [duplicate]

How do I get the path of the directory in which a Bash script is located, inside that script?
I want to use a Bash script as a launcher for another application. I want to change the working directory to the one where the Bash script is located, so I can operate on the files in that directory, like so:
$ ./application
#!/usr/bin/env bash
SCRIPT_DIR=$( cd -- "$( dirname -- "${BASH_SOURCE[0]}" )" &> /dev/null && pwd )
is a useful one-liner which will give you the full directory name of the script no matter where it is being called from.
It will work as long as the last component of the path used to find the script is not a symlink (directory links are OK). If you also want to resolve any links to the script itself, you need a multi-line solution:
#!/usr/bin/env bash
SOURCE=${BASH_SOURCE[0]}
while [ -L "$SOURCE" ]; do # resolve $SOURCE until the file is no longer a symlink
DIR=$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )
SOURCE=$(readlink "$SOURCE")
[[ $SOURCE != /* ]] && SOURCE=$DIR/$SOURCE # if $SOURCE was a relative symlink, we need to resolve it relative to the path where the symlink file was located
done
DIR=$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )
This last one will work with any combination of aliases, source, bash -c, symlinks, etc.
Beware: if you cd to a different directory before running this snippet, the result may be incorrect!
Also, watch out for $CDPATH gotchas, and stderr output side effects if the user has smartly overridden cd to redirect output to stderr instead (including escape sequences, such as when calling update_terminal_cwd >&2 on Mac). Adding >/dev/null 2>&1 at the end of your cd command will take care of both possibilities.
To understand how it works, try running this more verbose form:
#!/usr/bin/env bash
SOURCE=${BASH_SOURCE[0]}
while [ -L "$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 "$SOURCE" )" >/dev/null 2>&1 && pwd )
if [ "$DIR" != "$RDIR" ]; then
echo "DIR '$RDIR' resolves to '$DIR'"
fi
echo "DIR is '$DIR'"
And it will print something like:
SOURCE './scriptdir.sh' is a relative symlink to 'sym2/scriptdir.sh' (relative to '.')
SOURCE is './sym2/scriptdir.sh'
DIR './sym2' resolves to '/home/ubuntu/dotfiles/fo fo/real/real1/real2'
DIR is '/home/ubuntu/dotfiles/fo fo/real/real1/real2'
Use dirname "$0":
#!/usr/bin/env bash
echo "The script you are running has basename $( basename -- "$0"; ), dirname $( dirname -- "$0"; )";
echo "The present working directory is $( pwd; )";
Using pwd alone will not work if you are not running the script from the directory it is contained in.
[matt#server1 ~]$ pwd
/home/matt
[matt#server1 ~]$ ./test2.sh
The script you are running has basename test2.sh, dirname .
The present working directory is /home/matt
[matt#server1 ~]$ cd /tmp
[matt#server1 tmp]$ ~/test2.sh
The script you are running has basename test2.sh, dirname /home/matt
The present working directory is /tmp
The dirname command is the most basic, simply parsing the path up to the filename off of the $0 (script name) variable:
dirname -- "$0";
But, as matt b pointed out, the path returned is different depending on how the script is called. pwd doesn't do the job because that only tells you what the current directory is, not what directory the script resides in. Additionally, if a symbolic link to a script is executed, you're going to get a (probably relative) path to where the link resides, not the actual script.
Some others have mentioned the readlink command, but at its simplest, you can use:
dirname -- "$( readlink -f -- "$0"; )";
readlink will resolve the script path to an absolute path from the root of the filesystem. So, any paths containing single or double dots, tildes and/or symbolic links will be resolved to a full path.
Here's a script demonstrating each of these, whatdir.sh:
#!/usr/bin/env bash
echo "pwd: `pwd`"
echo "\$0: $0"
echo "basename: `basename -- "$0"`"
echo "dirname: `dirname -- "$0"`"
echo "dirname/readlink: $( dirname -- "$( readlink -f -- "$0"; )"; )"
Running this script in my home dir, using a relative path:
>>>$ ./whatdir.sh
pwd: /Users/phatblat
$0: ./whatdir.sh
basename: whatdir.sh
dirname: .
dirname/readlink: /Users/phatblat
Again, but using the full path to the script:
>>>$ /Users/phatblat/whatdir.sh
pwd: /Users/phatblat
$0: /Users/phatblat/whatdir.sh
basename: whatdir.sh
dirname: /Users/phatblat
dirname/readlink: /Users/phatblat
Now changing directories:
>>>$ cd /tmp
>>>$ ~/whatdir.sh
pwd: /tmp
$0: /Users/phatblat/whatdir.sh
basename: whatdir.sh
dirname: /Users/phatblat
dirname/readlink: /Users/phatblat
And finally using a symbolic link to execute the script:
>>>$ ln -s ~/whatdir.sh whatdirlink.sh
>>>$ ./whatdirlink.sh
pwd: /tmp
$0: ./whatdirlink.sh
basename: whatdirlink.sh
dirname: .
dirname/readlink: /Users/phatblat
There is however one case where this doesn't work, when the script is sourced (instead of executed) in bash:
>>>$ cd /tmp
>>>$ . ~/whatdir.sh
pwd: /tmp
$0: bash
basename: bash
dirname: .
dirname/readlink: /tmp
pushd . > '/dev/null';
SCRIPT_PATH="${BASH_SOURCE[0]:-$0}";
while [ -h "$SCRIPT_PATH" ];
do
cd "$( dirname -- "$SCRIPT_PATH"; )";
SCRIPT_PATH="$( readlink -f -- "$SCRIPT_PATH"; )";
done
cd "$( dirname -- "$SCRIPT_PATH"; )" > '/dev/null';
SCRIPT_PATH="$( pwd; )";
popd > '/dev/null';
It works for all versions, including
when called via multiple depth soft link,
when the file it
when script called by command "source" aka . (dot) operator.
when arg $0 is modified from caller.
"./script"
"/full/path/to/script"
"/some/path/../../another/path/script"
"./some/folder/script"
Alternatively, if the Bash script itself is a relative symlink you want to follow it and return the full path of the linked-to script:
pushd . > '/dev/null';
SCRIPT_PATH="${BASH_SOURCE[0]:-$0}";
while [ -h "$SCRIPT_PATH" ];
do
cd "$( dirname -- "$SCRIPT_PATH"; )";
SCRIPT_PATH="$( readlink -f -- "$SCRIPT_PATH"; )";
done
cd "$( dirname -- "$SCRIPT_PATH"; )" > '/dev/null';
SCRIPT_PATH="$( pwd; )";
popd > '/dev/null';
SCRIPT_PATH is given in full path, no matter how it is called.
Just make sure you locate this at start of the script.
You can use $BASH_SOURCE:
#!/usr/bin/env bash
scriptdir="$( dirname -- "$BASH_SOURCE"; )";
Note that you need to use #!/bin/bash and not #!/bin/sh since it's a Bash extension.
Here is an easy-to-remember script:
DIR="$( dirname -- "${BASH_SOURCE[0]}"; )"; # Get the directory name
DIR="$( realpath -e -- "$DIR"; )"; # Resolve its full path if need be
Short answer:
"`dirname -- "$0";`"
or (preferably):
"$( dirname -- "$0"; )"
This should do it:
DIR="$(dirname "$(realpath "$0")")"
This works with symlinks and spaces in path.
Please see the man pages for dirname and realpath.
Please add a comment on how to support MacOS. I'm sorry I can verify it.
pwd can be used to find the current working directory, and dirname to find the directory of a particular file (command that was run, is $0, so dirname $0 should give you the directory of the current script).
However, dirname gives precisely the directory portion of the filename, which more likely than not is going to be relative to the current working directory. If your script needs to change directory for some reason, then the output from dirname becomes meaningless.
I suggest the following:
#!/usr/bin/env bash
reldir="$( dirname -- "$0"; )";
cd "$reldir";
directory="$( pwd; )";
echo "Directory is ${directory}";
This way, you get an absolute, rather than a relative directory.
Since the script will be run in a separate Bash instance, there isn't any need to restore the working directory afterwards, but if you do want to change back in your script for some reason, you can easily assign the value of pwd to a variable before you change directory, for future use.
Although just
cd "$( dirname -- "$0"; )";
solves the specific scenario in the question, I find having the absolute path to more more useful generally.
SCRIPT_DIR=$( cd ${0%/*} && pwd -P )
I don't think this is as easy as others have made it out to be. pwd doesn't work, as the current directory is not necessarily the directory with the script. $0 doesn't always have the information either. Consider the following three ways to invoke a script:
./script
/usr/bin/script
script
In the first and third ways $0 doesn't have the full path information. In the second and third, pwd does not work. The only way to get the directory in the third way would be to run through the path and find the file with the correct match. Basically the code would have to redo what the OS does.
One way to do what you are asking would be to just hardcode the data in the /usr/share directory, and reference it by its full path. Data shoudn't be in the /usr/bin directory anyway, so this is probably the thing to do.
This gets the current working directory on Mac OS X v10.6.6 (Snow Leopard):
DIR=$(cd "$(dirname "$0")"; pwd)
$(dirname "$(readlink -f "$BASH_SOURCE")")
This is Linux specific, but you could use:
SELF=$(readlink /proc/$$/fd/255)
Here is a POSIX compliant one-liner:
SCRIPT_PATH=`dirname "$0"`; SCRIPT_PATH=`eval "cd \"$SCRIPT_PATH\" && pwd"`
# test
echo $SCRIPT_PATH
The shortest and most elegant way to do this is:
#!/bin/bash
DIRECTORY=$(cd `dirname $0` && pwd)
echo $DIRECTORY
This would work on all platforms and is super clean.
More details can be found in "Which directory is that bash script in?".
Summary:
FULL_PATH_TO_SCRIPT="$(realpath "${BASH_SOURCE[-1]}")"
# OR, if you do NOT need it to work for **sourced** scripts too:
# FULL_PATH_TO_SCRIPT="$(realpath "$0")"
# OR, depending on which path you want, in case of nested `source` calls
# FULL_PATH_TO_SCRIPT="$(realpath "${BASH_SOURCE[0]}")"
# OR, add `-s` to NOT expand symlinks in the path:
# FULL_PATH_TO_SCRIPT="$(realpath -s "${BASH_SOURCE[-1]}")"
SCRIPT_DIRECTORY="$(dirname "$FULL_PATH_TO_SCRIPT")"
SCRIPT_FILENAME="$(basename "$FULL_PATH_TO_SCRIPT")"
Details:
How to obtain the full file path, full directory, and base filename of any script being run OR sourced...
...even when the called script is called from within another bash function or script, or when nested sourcing is being used!
For many cases, all you need to acquire is the full path to the script you just called. This can be easily accomplished using realpath. Note that realpath is part of GNU coreutils. If you don't have it already installed (it comes default on Ubuntu), you can install it with sudo apt update && sudo apt install coreutils.
get_script_path.sh (for the latest version of this script, see get_script_path.sh in my eRCaGuy_hello_world repo):
#!/bin/bash
# A. Obtain the full path, and expand (walk down) symbolic links
# A.1. `"$0"` works only if the file is **run**, but NOT if it is **sourced**.
# FULL_PATH_TO_SCRIPT="$(realpath "$0")"
# A.2. `"${BASH_SOURCE[-1]}"` works whether the file is sourced OR run, and even
# if the script is called from within another bash function!
# NB: if `"${BASH_SOURCE[-1]}"` doesn't give you quite what you want, use
# `"${BASH_SOURCE[0]}"` instead in order to get the first element from the array.
FULL_PATH_TO_SCRIPT="$(realpath "${BASH_SOURCE[-1]}")"
# B.1. `"$0"` works only if the file is **run**, but NOT if it is **sourced**.
# FULL_PATH_TO_SCRIPT_KEEP_SYMLINKS="$(realpath -s "$0")"
# B.2. `"${BASH_SOURCE[-1]}"` works whether the file is sourced OR run, and even
# if the script is called from within another bash function!
# NB: if `"${BASH_SOURCE[-1]}"` doesn't give you quite what you want, use
# `"${BASH_SOURCE[0]}"` instead in order to get the first element from the array.
FULL_PATH_TO_SCRIPT_KEEP_SYMLINKS="$(realpath -s "${BASH_SOURCE[-1]}")"
# You can then also get the full path to the directory, and the base
# filename, like this:
SCRIPT_DIRECTORY="$(dirname "$FULL_PATH_TO_SCRIPT")"
SCRIPT_FILENAME="$(basename "$FULL_PATH_TO_SCRIPT")"
# Now print it all out
echo "FULL_PATH_TO_SCRIPT = \"$FULL_PATH_TO_SCRIPT\""
echo "SCRIPT_DIRECTORY = \"$SCRIPT_DIRECTORY\""
echo "SCRIPT_FILENAME = \"$SCRIPT_FILENAME\""
IMPORTANT note on nested source calls: if "${BASH_SOURCE[-1]}" above doesn't give you quite what you want, try using "${BASH_SOURCE[0]}" instead. The first (0) index gives you the first entry in the array, and the last (-1) index gives you the last last entry in the array. Depending on what it is you're after, you may actually want the first entry. I discovered this to be the case when I sourced ~/.bashrc with . ~/.bashrc, which sourced ~/.bash_aliases with . ~/.bash_aliases, and I wanted the realpath (with expanded symlinks) to the ~/.bash_aliases file, NOT to the ~/.bashrc file. Since these are nested source calls, using "${BASH_SOURCE[0]}" gave me what I wanted: the expanded path to ~/.bash_aliases! Using "${BASH_SOURCE[-1]}", however, gave me what I did not want: the expanded path to ~/.bashrc.
Example command and output:
Running the script:
~/GS/dev/eRCaGuy_hello_world/bash$ ./get_script_path.sh
FULL_PATH_TO_SCRIPT = "/home/gabriel/GS/dev/eRCaGuy_hello_world/bash/get_script_path.sh"
SCRIPT_DIRECTORY = "/home/gabriel/GS/dev/eRCaGuy_hello_world/bash"
SCRIPT_FILENAME = "get_script_path.sh"
Sourcing the script with . get_script_path.sh or source get_script_path.sh (the result is the exact same as above because I used "${BASH_SOURCE[-1]}" in the script instead of "$0"):
~/GS/dev/eRCaGuy_hello_world/bash$ . get_script_path.sh
FULL_PATH_TO_SCRIPT = "/home/gabriel/GS/dev/eRCaGuy_hello_world/bash/get_script_path.sh"
SCRIPT_DIRECTORY = "/home/gabriel/GS/dev/eRCaGuy_hello_world/bash"
SCRIPT_FILENAME = "get_script_path.sh"
If you use "$0" in the script instead of "${BASH_SOURCE[-1]}", you'll get the same output as above when running the script, but this undesired output instead when sourcing the script:
~/GS/dev/eRCaGuy_hello_world/bash$ . get_script_path.sh
FULL_PATH_TO_SCRIPT = "/bin/bash"
SCRIPT_DIRECTORY = "/bin"
SCRIPT_FILENAME = "bash"
And, apparently if you use "$BASH_SOURCE" instead of "${BASH_SOURCE[-1]}", it will not work if the script is called from within another bash function. So, using "${BASH_SOURCE[-1]}" is therefore the best way to do it, as it solves both of these problems! See the references below.
Difference between realpath and realpath -s:
Note that realpath also successfully walks down symbolic links to determine and point to their targets rather than pointing to the symbolic link. If you do NOT want this behavior (sometimes I don't), then add -s to the realpath command above, making that line look like this instead:
# Obtain the full path, but do NOT expand (walk down) symbolic links; in
# other words: **keep** the symlinks as part of the path!
FULL_PATH_TO_SCRIPT="$(realpath -s "${BASH_SOURCE[-1]}")"
This way, symbolic links are NOT expanded. Rather, they are left as-is, as symbolic links in the full path.
The code above is now part of my eRCaGuy_hello_world repo in this file here: bash/get_script_path.sh. Reference and run this file for full examples both with and withOUT symlinks in the paths. See the bottom of the file for example output in both cases.
References:
How to retrieve absolute path given relative
taught me about the BASH_SOURCE variable: Unix & Linux: determining path to sourced shell script
taught me that BASH_SOURCE is actually an array, and we want the last element from it for it to work as expected inside a function (hence why I used "${BASH_SOURCE[-1]}" in my code here): Unix & Linux: determining path to sourced shell script
man bash --> search for BASH_SOURCE:
BASH_SOURCE
An array variable whose members are the source filenames where the corresponding shell function names in the FUNCNAME array variable are defined. The shell function ${FUNCNAME[$i]} is defined in the file ${BASH_SOURCE[$i]} and called from ${BASH_SOURCE[$i+1]}.
See also:
[my answer] Unix & Linux: determining path to sourced shell script
#!/bin/sh
PRG="$0"
# need this for relative symlinks
while [ -h "$PRG" ] ; do
PRG=`readlink "$PRG"`
done
scriptdir=`dirname "$PRG"`
Here is the simple, correct way:
actual_path=$(readlink -f "${BASH_SOURCE[0]}")
script_dir=$(dirname "$actual_path")
Explanation:
${BASH_SOURCE[0]} - the full path to the script. The value of this will be correct even when the script is being sourced, e.g. source <(echo 'echo $0') prints bash, while replacing it with ${BASH_SOURCE[0]} will print the full path of the script. (Of course, this assumes you're OK taking a dependency on Bash.)
readlink -f - Recursively resolves any symlinks in the specified path. This is a GNU extension, and not available on (for example) BSD systems. If you're running a Mac, you can use Homebrew to install GNU coreutils and supplant this with greadlink -f.
And of course dirname gets the parent directory of the path.
I tried all of these and none worked. One was very close, but it had a tiny bug that broke it badly; they forgot to wrap the path in quotation marks.
Also a lot of people assume you're running the script from a shell, so they forget when you open a new script it defaults to your home.
Try this directory on for size:
/var/No one/Thought/About Spaces Being/In a Directory/Name/And Here's your file.text
This gets it right regardless how or where you run it:
#!/bin/bash
echo "pwd: `pwd`"
echo "\$0: $0"
echo "basename: `basename "$0"`"
echo "dirname: `dirname "$0"`"
So to make it actually useful, here's how to change to the directory of the running script:
cd "`dirname "$0"`"
This is a slight revision to the solution e-satis and 3bcdnlklvc04a pointed out in their answer:
SCRIPT_DIR=''
pushd "$(dirname "$(readlink -f "$BASH_SOURCE")")" > /dev/null && {
SCRIPT_DIR="$PWD"
popd > /dev/null
}
This should still work in all the cases they listed.
This will prevent popd after a failed pushd. Thanks to konsolebox.
Try using:
real=$(realpath "$(dirname "$0")")
I would use something like this:
# Retrieve the full pathname of the called script
scriptPath=$(which $0)
# Check whether the path is a link or not
if [ -L $scriptPath ]; then
# It is a link then retrieve the target path and get the directory name
sourceDir=$(dirname $(readlink -f $scriptPath))
else
# Otherwise just get the directory name of the script path
sourceDir=$(dirname $scriptPath)
fi
For systems having GNU coreutils readlink (for example, Linux):
$(readlink -f "$(dirname "$0")")
There's no need to use BASH_SOURCE when $0 contains the script filename.
$_ is worth mentioning as an alternative to $0. If you're running a script from Bash, the accepted answer can be shortened to:
DIR="$( dirname "$_" )"
Note that this has to be the first statement in your script.
These are short ways to get script information:
Folders and files:
Script: "/tmp/src dir/test.sh"
Calling folder: "/tmp/src dir/other"
Using these commands:
echo Script-Dir : `dirname "$(realpath $0)"`
echo Script-Dir : $( cd ${0%/*} && pwd -P )
echo Script-Dir : $(dirname "$(readlink -f "$0")")
echo
echo Script-Name : `basename "$(realpath $0)"`
echo Script-Name : `basename $0`
echo
echo Script-Dir-Relative : `dirname "$BASH_SOURCE"`
echo Script-Dir-Relative : `dirname $0`
echo
echo Calling-Dir : `pwd`
And I got this output:
Script-Dir : /tmp/src dir
Script-Dir : /tmp/src dir
Script-Dir : /tmp/src dir
Script-Name : test.sh
Script-Name : test.sh
Script-Dir-Relative : ..
Script-Dir-Relative : ..
Calling-Dir : /tmp/src dir/other
Also see: https://pastebin.com/J8KjxrPF
This works in Bash 3.2:
path="$( dirname "$( which "$0" )" )"
If you have a ~/bin directory in your $PATH, you have A inside this directory. It sources the script ~/bin/lib/B. You know where the included script is relative to the original one, in the lib subdirectory, but not where it is relative to the user's current directory.
This is solved by the following (inside A):
source "$( dirname "$( which "$0" )" )/lib/B"
It doesn't matter where the user is or how he/she calls the script. This will always work.
I've compared many of the answers given, and came up with some more compact solutions. These seem to handle all of the crazy edge cases that arise from your favorite combination of:
Absolute paths or relative paths
File and directory soft links
Invocation as script, bash script, bash -c script, source script, or . script
Spaces, tabs, newlines, Unicode, etc. in directories and/or filename
Filenames beginning with a hyphen
If you're running from Linux, it seems that using the proc handle is the best solution to locate the fully resolved source of the currently running script (in an interactive session, the link points to the respective /dev/pts/X):
resolved="$(readlink /proc/$$/fd/255 && echo X)" && resolved="${resolved%$'\nX'}"
This has a small bit of ugliness to it, but the fix is compact and easy to understand. We aren't using bash primitives only, but I'm okay with that because readlink simplifies the task considerably. The echo X adds an X to the end of the variable string so that any trailing whitespace in the filename doesn't get eaten, and the parameter substitution ${VAR%X} at the end of the line gets rid of the X. Because readlink adds a newline of its own (which would normally be eaten in the command substitution if not for our previous trickery), we have to get rid of that, too. This is most easily accomplished using the $'' quoting scheme, which lets us use escape sequences such as \n to represent newlines (this is also how you can easily make deviously named directories and files).
The above should cover your needs for locating the currently running script on Linux, but if you don't have the proc filesystem at your disposal, or if you're trying to locate the fully resolved path of some other file, then maybe you'll find the below code helpful. It's only a slight modification from the above one-liner. If you're playing around with strange directory/filenames, checking the output with both ls and readlink is informative, as ls will output "simplified" paths, substituting ? for things like newlines.
absolute_path=$(readlink -e -- "${BASH_SOURCE[0]}" && echo x) && absolute_path=${absolute_path%?x}
dir=$(dirname -- "$absolute_path" && echo x) && dir=${dir%?x}
file=$(basename -- "$absolute_path" && echo x) && file=${file%?x}
ls -l -- "$dir/$file"
printf '$absolute_path: "%s"\n' "$absolute_path"
I believe I've got this one. I'm late to the party, but I think some will appreciate it being here if they come across this thread. The comments should explain:
#!/bin/sh # dash bash ksh # !zsh (issues). G. Nixon, 12/2013. Public domain.
## 'linkread' or 'fullpath' or (you choose) is a little tool to recursively
## dereference symbolic links (ala 'readlink') until the originating file
## is found. This is effectively the same function provided in stdlib.h as
## 'realpath' and on the command line in GNU 'readlink -f'.
## Neither of these tools, however, are particularly accessible on the many
## systems that do not have the GNU implementation of readlink, nor ship
## with a system compiler (not to mention the requisite knowledge of C).
## This script is written with portability and (to the extent possible, speed)
## in mind, hence the use of printf for echo and case statements where they
## can be substituded for test, though I've had to scale back a bit on that.
## It is (to the best of my knowledge) written in standard POSIX shell, and
## has been tested with bash-as-bin-sh, dash, and ksh93. zsh seems to have
## issues with it, though I'm not sure why; so probably best to avoid for now.
## Particularly useful (in fact, the reason I wrote this) is the fact that
## it can be used within a shell script to find the path of the script itself.
## (I am sure the shell knows this already; but most likely for the sake of
## security it is not made readily available. The implementation of "$0"
## specificies that the $0 must be the location of **last** symbolic link in
## a chain, or wherever it resides in the path.) This can be used for some
## ...interesting things, like self-duplicating and self-modifiying scripts.
## Currently supported are three errors: whether the file specified exists
## (ala ENOENT), whether its target exists/is accessible; and the special
## case of when a sybolic link references itself "foo -> foo": a common error
## for beginners, since 'ln' does not produce an error if the order of link
## and target are reversed on the command line. (See POSIX signal ELOOP.)
## It would probably be rather simple to write to use this as a basis for
## a pure shell implementation of the 'symlinks' util included with Linux.
## As an aside, the amount of code below **completely** belies the amount
## effort it took to get this right -- but I guess that's coding for you.
##===-------------------------------------------------------------------===##
for argv; do :; done # Last parameter on command line, for options parsing.
## Error messages. Use functions so that we can sub in when the error occurs.
recurses(){ printf "Self-referential:\n\t$argv ->\n\t$argv\n" ;}
dangling(){ printf "Broken symlink:\n\t$argv ->\n\t"$(readlink "$argv")"\n" ;}
errnoent(){ printf "No such file: "$#"\n" ;} # Borrow a horrible signal name.
# Probably best not to install as 'pathfull', if you can avoid it.
pathfull(){ cd "$(dirname "$#")"; link="$(readlink "$(basename "$#")")"
## 'test and 'ls' report different status for bad symlinks, so we use this.
if [ ! -e "$#" ]; then if $(ls -d "$#" 2>/dev/null) 2>/dev/null; then
errnoent 1>&2; exit 1; elif [ ! -e "$#" -a "$link" = "$#" ]; then
recurses 1>&2; exit 1; elif [ ! -e "$#" ] && [ ! -z "$link" ]; then
dangling 1>&2; exit 1; fi
fi
## Not a link, but there might be one in the path, so 'cd' and 'pwd'.
if [ -z "$link" ]; then if [ "$(dirname "$#" | cut -c1)" = '/' ]; then
printf "$#\n"; exit 0; else printf "$(pwd)/$(basename "$#")\n"; fi; exit 0
fi
## Walk the symlinks back to the origin. Calls itself recursivly as needed.
while [ "$link" ]; do
cd "$(dirname "$link")"; newlink="$(readlink "$(basename "$link")")"
case "$newlink" in
"$link") dangling 1>&2 && exit 1 ;;
'') printf "$(pwd)/$(basename "$link")\n"; exit 0 ;;
*) link="$newlink" && pathfull "$link" ;;
esac
done
printf "$(pwd)/$(basename "$newlink")\n"
}
## Demo. Install somewhere deep in the filesystem, then symlink somewhere
## else, symlink again (maybe with a different name) elsewhere, and link
## back into the directory you started in (or something.) The absolute path
## of the script will always be reported in the usage, along with "$0".
if [ -z "$argv" ]; then scriptname="$(pathfull "$0")"
# Yay ANSI l33t codes! Fancy.
printf "\n\033[3mfrom/as: \033[4m$0\033[0m\n\n\033[1mUSAGE:\033[0m "
printf "\033[4m$scriptname\033[24m [ link | file | dir ]\n\n "
printf "Recursive readlink for the authoritative file, symlink after "
printf "symlink.\n\n\n \033[4m$scriptname\033[24m\n\n "
printf " From within an invocation of a script, locate the script's "
printf "own file\n (no matter where it has been linked or "
printf "from where it is being called).\n\n"
else pathfull "$#"
fi
Try the following cross-compatible solution:
CWD="$(cd -P -- "$(dirname -- "${BASH_SOURCE[0]}")" && pwd -P)"
As the commands such as realpath or readlink could be not available (depending on the operating system).
Note: In Bash, it's recommended to use ${BASH_SOURCE[0]} instead of $0, otherwise path can break when sourcing the file (source/.).
Alternatively you can try the following function in Bash:
realpath () {
[[ $1 = /* ]] && echo "$1" || echo "$PWD/${1#./}"
}
This function takes one argument. If argument has already absolute path, print it as it is, otherwise print $PWD variable + filename argument (without ./ prefix).
Related:
How can I set the current working directory to the directory of the script in Bash?
Bash script absolute path with OS X
Reliable way for a Bash script to get the full path to itself

How can I get the path to the currently running shell script (cshell linux)

Linux/CSHELL:
How can I get the path to a tcsh script that is being sourced?
This is not $PWD. Rather, the path to the script (file).
Versions/specifics:
% echo $SHELL $PLATFORM
/tool/pandora/bin/tcsh linux_3.10.0_64
ls -l /tool/pandora/bin/tcsh
lrwxrwxrwx 1 pandora pandora 33 Oct 20 2016 /tool/pandora/bin/tcsh -> ../.package/tcsh-6.19.00/bin/tcsh
There are lots of notes in here for this type of thing. But I didn't see one for linux c-shell.
Thanks in Advance !
Example:
I have a file /home/myid/my_cshell.csh...
set a = `readlink -f ${0}`
echo $a
echo ${0}
source /home/myid/my_cshell.csh
/tool/pandora/.package/tcsh-6.19.00/bin/tcsh
/tool/pandora/bin/tcsh
This is giving me the path to tcsh, not /home/myid/my_cshell.csh
The script above works fine if I tcsh it, but not source it.
I need the path to the file being sourced.
assuming you have readlink available:
readlink -f "${0}"
should work. ${0} will give you the relative path, and readlink -f will resolve it to the absolute path.
There must already be some way to reference the file... If not, how could it be sourced in the first place? So I suppose the problem is that ${0} referecnes the parent's file system location, and not the sourced file. In that case, readlink -f <file> should still work, but maybe try calling it first, and then pass the result when you run source. Like this:
main.csh:
#!/bin/csh
set p=`readlink -f source.csh`
source source.csh "${p}"
source.csh:
#!/bin/csh
set pmain=`readlink -f "${0}"`
echo "my path is ${1}. main.csh's path is ${pmain}"

Bash script: Can't set a variable using which

In my terminal,
prog="cat"
name=$(which $prog)
echo $name
prints /bin/cat
But in my script:
pro="$1"
prog=$(which $pro)
echo "pro is $pro"
echo "prog is "$prog""
running scriptname cat prints
pro is cat
prog is
How do I make which work? it should print prog is /bin/cat
which(1) is an external program used to search PATH for an executable. It behaves differently on different systems and you can't rely on a useful exit code; use (from most to least portable) command -v or type -P (to find the path) or hash (to check) instead.
try printf '%s\n' "$PATH" inside your script as well as outside of it. maybe the command you're looking for is not in the PATH used in the script?
That is almost certainly the cause.

bash - how to pipe result from the which command to cd

How could I pipe the result from a which command to cd?
This is what I am trying to do:
which oracle | cd
cd < which oracle
But none of them works.
Is there a way to achieve this (rather than copy/paste of course)?
Edit : on second thought, this command would fail, because the destination file is NOT a folder/directory.
So I am thinking and working out a better way to get rid of the trailing "/oracle" part now (sed or awk, or even Perl) :)
Edit :
Okay that's what I've got in the end:
cd `which oracle | sed 's/\/oracle//g'`
You use pipe in cases where the command expects parameters from the standard input. ( More on this ).
With cd command that is not the case. The directory is the command argument. In such case, you can use command substitution. Use backticks or $(...) to evaluate the command, store it into variable..
path=`which oracle`
echo $path # just for debug
cd $path
although it can be done in a much simpler way:
cd `which oracle`
or if your path has special characters
cd "`which oracle`"
or
cd $(which oracle)
which is equivalent to backtick notation, but is recommended (backticks can be confused with apostrophes)
.. but it looks like you want:
cd $(dirname $(which oracle))
(which shows you that you can use nesting easily)
$(...) (as well as backticks) work also in double-quoted strings, which helps when the result may eventually contain spaces..
cd "$(dirname "$(which oracle)")"
(Note that both outputs require a set of double quotes.)
With dirname to get the directory:
cd $(which oracle | xargs dirname)
EDIT: beware of paths containing spaces, see #anishpatel comment below
cd `which oracle`
Note those are backticks (generally the key to the left of 1 on a US keyboard)
OK, here a solution that uses correct quoting:
cd "$(dirname "$(which oracle)")"
Avoid backticks, they are less readable, and always quote process substitutions.
You don't need a pipe, you can do what you want using Bash parameter expansion!
Further tip: use "type -P" instead of the external "which" command if you are using Bash.
# test
touch /ls
chmod +x /ls
cmd='ls'
PATH=/:$PATH
if cmdpath="$(type -P "$cmd")" && cmdpath="${cmdpath%/*}" ; then
cd "${cmdpath:-/}" || { echo "Could not cd to: ${cmdpath:-/}"; exit 1; }
else
echo "No such program in PATH search directories: ${cmd}"
exit 1
fi
besides good answer above, one thing needs to mention is that cd is a shell builtin, which run in the same process other than new process like ls which is a command.
https://unix.stackexchange.com/questions/50022/why-cant-i-redirect-a-path-name-output-from-one-command-to-cd
http://en.wikipedia.org/wiki/Shell_builtin
In response to your edited question, you can strip off the name of the command using dirname:
cd $(dirname `which oracle`)

How to get script file path inside script itself when called through sym link

When I need to get path to the script file inside script itself I use something like this:
`dirname $0`
that works file until I call the script through sym link to it. In that case above code prints the location of the link instead the original file.
Is there a way to get the path of the original script file, not the link?
Thanks,
Mike
if [ -L $0 ] ; then
DIR=$(dirname $(readlink -f $0)) ;
else
DIR=$(dirname $0) ;
fi ;
You really need to read the following BashFAQ article:
BashFAQ/028 - How do I determine the location of my script? I want to read some config files from the same place.
The truth is, the $0 hacks are NOT reliable, and they will fail whenever applications are started with a zeroth argument that is not the application's path. For example, login(1) will put a - in front of your application name in $0, causing breakage, and whenever you invoke an application that's in PATH $0 won't contain the path to your application but just the filename, which means you can't extract any location information out of it. There are a LOT more ways in which this can fail.
Bottom line: Don't rely on the hack, the truth is you cannot figure out where your script is, so use a configuration option to set the directory of your config files or whatever you need the script's path for.
Use readlink for that:
readlink -f "$0"
Dir=$(dirname $(readlink -f "$0"))
I suggest this one:
dirname $(realpath $0)
example:
skaerst#test01-centos7:~/Documents> ../test/path.sh
dirname realpath: /home/skaerst/test
skaerst#test01-centos7:~/Documents> cat ../test/path.sh
#!/bin/bash
echo "dirname realpath: $(dirname $(realpath $0))"
works with symlinks too because realpath uses -P by default:
skaerst#test01-centos7:~/Documents> ln -s ../test/path.sh p
skaerst#test01-centos7:~/Documents> ./p
dirname realpath: /home/skaerst/test
realpath is available with coreutils >8.5, I guess
skaerst#test01-centos7:~/Documents> rpm -qf $(which realpath) $(which dirname)
coreutils-8.22-15.el7_2.1.x86_64
coreutils-8.22-15.el7_2.1.x86_64
hth
Regards
Stefan

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