Get The Name of The Directory Where a Script Is Executed

How do I get the directory where a Bash script is located from within the script itself?

#!/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'

Get the name of the directory where a script is executed

Get the real path to your script

if [ -L $0 ] ; then
    ME=$(readlink $0)
else
    ME=$0
fi
DIR=$(dirname $ME)

get script directory name - Python

import os
os.path.basename(os.path.dirname(os.path.realpath(__file__)))

Broken down:

currentFile = __file__  # May be 'my_script', or './my_script' or
                        # '/home/user/test/my_script.py' depending on exactly how
                        # the script was run/loaded.
realPath = os.path.realpath(currentFile)  # /home/user/test/my_script.py
dirPath = os.path.dirname(realPath)  # /home/user/test
dirName = os.path.basename(dirPath) # test

What's the best way to determine the location of the current PowerShell script?

PowerShell 3+

# This is an automatic variable set to the current file's/module's directory
$PSScriptRoot

PowerShell 2

Prior to PowerShell 3, there was not a better way than querying the
MyInvocation.MyCommand.Definition property for general scripts. I had the following line at the top of essentially every PowerShell script I had:

$scriptPath = split-path -parent $MyInvocation.MyCommand.Definition

How can I find script's directory?

You need to call os.path.realpath on __file__, so that when __file__ is a filename without the path you still get the dir path:

import os
print(os.path.dirname(os.path.realpath(__file__)))

Unix shell script find out which directory the script file resides?

In Bash, you should get what you need like this:

#!/usr/bin/env bash

BASEDIR=$(dirname "$0")
echo "$BASEDIR"

bash - get directory a script is run from

The $PWD variable is probably what you need.

$ cat >/tmp/pwd.bash <<'END'
#!/bin/bash
echo "\$0=$0"
echo "\$PWD=$PWD"
END

$ chmod u+x /tmp/pwd.bash

$ pwd
/home/jackman

$ /tmp/pwd.bash
$0=/tmp/pwd.bash
$PWD=/home/jackman

How do I know the script file name in a Bash script?

me=`basename "$0"`

For reading through a symlink¹, which is usually not what you want (you usually don't want to confuse the user this way), try:

me="$(basename "$(test -L "$0" && readlink "$0" || echo "$0")")"

IMO, that'll produce confusing output. "I ran foo.sh, but it's saying I'm running bar.sh!? Must be a bug!" Besides, one of the purposes of having differently-named symlinks is to provide different functionality based on the name it's called as (think gzip and gunzip on some platforms).

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¹ That is, to resolve symlinks such that when the user executes foo.sh which is actually a symlink to bar.sh, you wish to use the resolved name bar.sh rather than foo.sh.

How do I get the directory of the PowerShell script I execute?

PowerShell 3 has the $PSScriptRoot automatic variable:

Contains the directory from which a script is being run.

In Windows PowerShell 2.0, this variable is valid only in script modules (.psm1). Beginning in Windows PowerShell 3.0, it is valid in all scripts.

Don't be fooled by the poor wording. PSScriptRoot is the directory of the current file.

In PowerShell 2, you can calculate the value of $PSScriptRoot yourself:

# PowerShell v2
$PSScriptRoot = Split-Path -Parent -Path $MyInvocation.MyCommand.Definition

How do I find directory of the Python running script from inside the script?

To get the directory that contains the module you are running:

import os
path = os.path.dirname(os.path.realpath(__file__))

Or if you want the directory from which the script was invoked:

import os
path = os.getcwd()

From the docs:

__file__ is the pathname of the file from which the module was loaded, if it was loaded from a file.

Depending on how the script is called, this may be a relative path from os.getcwd(), so os.path.realpath(__file__) will convert this to an absolute path (or do nothing is the __file__ is already an absolute path). os.path.dirname() will then return the full directory by stripping off the filename.

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