Unix shell script find out which directory the scr

问题:

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?

回答1:

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

#!/usr/bin/env bash

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

回答2:

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

回答3:

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

回答4:

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.

回答5:

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}"

回答6:

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.

回答7:

The best answer for this question was answered here:
Getting the source directory of a Bash script from within

And 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'"

回答8:

cd $(dirname $(readlink -f $0))

回答9:

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/

回答10:

BASE_DIR="$(cd "$(dirname "$0")"; pwd)";
echo "BASE_DIR => $BASE_DIR"

回答11:

Inspired by blueyed’s answer

read < <(readlink -f $0 | xargs dirname)
cd $REPLY

回答12:

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)"

回答13:

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

回答14:

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.

回答15:

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).