I have a file .aliases that I point to in my .zshrc file
DIR="$( cd "$( dirname "${BASH_SOURCE[0]}" )" && pwd )"
if [ -f $DIR"/.aliases" ]; then
. $DIR"/.aliases"
fi
This code is the same in my .bashrc file (in the process of switching over to zsh). The aliases work fine when I open a new terminal - however when I change directory and open tmux, the aliases break. At first I thought this had something to do with whether the directory is the correct one - however, when I switch to bash, it works! So I am very confused as to what is causing this issue.
BASH_SOURCE cannot be used in .zshrc, because it is a bash-specific variable that isn't defined in zsh. You'll have to replace it with its zsh equivalent, found here.
Related
I wonder if there's a good way to source another shell script without $(basename "$0") solution, since sometimes "$0" is not set.
For exmaple "$0" would be "-sh" for login shell.
There's same problem with regard to cwd solution.
You could use this expression to find out the directory of the parent Bash script:
dir="$( cd "$( dirname "${BASH_SOURCE[0]}" )" && pwd )"
Then you can call your script like this:
cd $dir; source ./other_script
or
source $dir/other_script
See this related post: Getting the source directory of a Bash script from within
In bash i get the executing script's parent folder name by this line
SCRIPT_PARENT=`readlink -f ${BASH_SOURCE%/*}/..`
Is there any way to achieve this in zsh in a way that works both in zsh and bash?
Assume i have got a file /some/folder/rootfolder/subfolder/script with the contents:
echo `magic-i-am-looking-for`
I want it to behave this way:
$ cd /some/other/folder
$ . /some/folder/rootfolder/subfolder/script
/some/folder/rootfolder
$ . ../../folder/rootfolder/subfolder/script
/some/folder/rootfolder
$ cd /some/folder/rootfolder
$ . subfolder/script
/some/folder/rootfolder
$ cd subfolder
$ . script
/some/folder/rootfolder
This should work in bash and zsh. My first implements this behavior, but does due to $BASH_SOURCE not work in zsh.
So basically its:
Is there a way to emulate $BASH_SOURCE in zsh, that also works in bash?
I now realized that $0 in zsh behaves like $BASH_SOURCE in bash. So using $BASH_SOURCE when available and falling back to $0 solves my problem:
${BASH_SOURCE:-$0}
There is a little zsh edge case left, when sourcing from $PATH like:
zsh> cat ../script
echo \$0: $0
echo \$BASH_SOURCE: $BASH_SOURCE
echo '${BASH_SOURCE:-$0}:' ${BASH_SOURCE:-$0}
zsh> . script
$0: script
$BASH_SOURCE:
${BASH_SOURCE:-$0}: script
bash> . script
$0: bash
$BASH_SOURCE: /home/me/script
${BASH_SOURCE:-$0}: /home/me/script
I could do a which script but this would not play nice with other cases
While it would be easy to do this in zsh, it is just as easy to use pure bash which is able to be evaluated in zsh. If you cannot use any command that may or may not be on your path, then you can only use variable alteration to achieve what you want:
SCRIPT_SOURCE=${0%/*}
This is likely to be a relative path. If you really want the full path then you will have to resort to an external command (you could implement it yourself, but it would be a lot of work to avoid using a very available command):
SCRIPT_SOURCE=$(/bin/readlink -f ${0%/*})
This doesn't depend on your $PATH, it just depends on /bin/readlink being present. Which it almost certainly is.
Now, you wanted this to be a sourced file. This is fine, as you can just export any variable you set, however if you execute the above then $0 will be the location of the sourced file and not the location of the calling script.
This just means you need to set a variable to hold the $0 value which the sourced script knows about. For example:
The script you will source:
echo ${LOCATION%/*}
The script that sources that script:
LOCATION=$0
<source script here>
But given that the ${0%/*} expansion is so compact, you could just use that in place of the script.
Because you were able to run the command from your $PATH I'll do something like that:
SCRIPT_PARENT=$(readlink -f "$(which $0)/..")
Is that your desired output?
I am looking for a simple solution to retrieve the absolute path of the current script. It needs to be platform independent (I want it to work on linux, freebsd, macos and without bash).
"readlink -f $0" works on linux but not on freebsd and macos: readlink
doesn't have the "-f" option.
"realpath $0" works on freebsd and linux but not on macos: I don't have this command.
EDIT :
Solution for retrieve the path of the repository of the script :
DIR="$( cd "$( dirname "$0" )" && pwd )" (source : Getting the source directory of a Bash script from within )
#!/bin/sh
self=$(
self=${0}
while [ -L "${self}" ]
do
cd "${self%/*}"
self=$(readlink "${self}")
done
cd "${self%/*}"
echo "$(pwd -P)/${self##*/}"
)
echo "${self}"
It's «mostly portable». Pattern substitution and pwd -P is POSIX, and the latter is usually a shell built-in. readlink is pretty common but it's not in POSIX.
And I don't think there is a simpler mostly-portable way. If you really need something like that, I'd suggest you rather try to get realpath installed on all your systems.
For zsh scripts, FWIW:
#! /bin/zsh -
fullpath=$0:A
I've somehow managed to screw up bash while fiddling with the $PATH variable in my bash_profile (I think...). All I did, as far as I can remember, was add a directory to the $PATH variable. Please HELP!
Here's what I get when I cd into various directories
my-MacBook-Pro:~ myuser$ cd .rvm
-bash: dirname: command not found
-bash: find: command not found
my-MacBook-Pro:.rvm myuser$ cd
-bash: find: command not found
And here's what happens when I try to get into my .bash_profile to undo whatever it is that I did...
my-MacBook-Pro:~ myuser$ emacs .bash_profile
-bash: emacs: command not found
my-MacBook-Pro:~ myuser$ sudo emacs .bash_profile
-bash: sudo: command not found
Any help would be massively appreciated. I'm completely screwed until I can get bash working normally again!
/usr/bin/emacs .bash_profile or similar should work when the PATH is broken.
The $PATH variable tells the shell where to look for commands. If you just bypass that by telling it the full path, it should work. Try /usr/bin/emacs .bash_profile.
When you do a cd, you're getting a bunch of other things. Since you're using BASH there are are two possible issues:
You have PROMPT_COMMAND defined. Try to undefining it:
$ unset PROMPT_COMMAND
There's an alias of the cd command: This was quite common in Kornshell where you don't have the nice backslashed characters you could put into your prompt string. If you wanted your prompt to have the name of your directory in it.
You had to do something like this:
function _cd
{
logname="$(logname)"
hostname="$(hostname)"
directory="$1"
pattern="$2"
if [ "$pattern" ] #This is a substitution!
then
\cd "$directory" "$pattern"
elif [ "$directory" ]
then
\cd "$directory"
else
\cd
fi
directory=$PWD
shortName=${directory#$HOME}
if [ "$shortName" = "" ]
then
prompt="~$logname"
elif [ "$shortName" = "$directory" ]
then
prompt="$directory"
else
prompt="~$shortName"
fi
title="$logname#$hostname:$prompt"
PS1="$title
$ "
}
alias cd="_cd"
Ugly isn't it? You don't have to go through all of that for BASH, but this does work in BASH too, and I've seen places where this was done either out of ignorance of inertia.
Try this:
$ type cd
You'll either get
$type cd
cd is a shell builtin
or you'll get
$ type cd
cd is an alias for ....
As for your updating of $PATH, you probably forgot to put $PATH back in the new definition, or quotation marks because someone has a directory name with a space in it. Your PATH setting should look like this:
PATH="/my/directory:$PATH"
Some people say it should be:
PATH="$PATH:/my/directory"
I guess, that you have defined $PROMPT_COMMAND (maybe in .bashrc) in a way that uses dirname and find.
That would explain the behavior of cd.
The find command is by default in /usr/bin/find. Thus, you can use it to find the locations of your imprtant commands and reconstruct you path information.
I'm writing a Bash script. I need the current working directory to always be the directory that the script is located in.
The default behavior is that the current working directory in the script is that of the shell from which I run it, but I do not want this behavior.
#!/bin/bash
cd "$(dirname "$0")"
The following also works:
cd "${0%/*}"
The syntax is thoroughly described in this StackOverflow answer.
Try the following simple one-liners:
For all UNIX/OSX/Linux
dir="$(cd -P -- "$(dirname -- "$0")" && pwd -P)"
Bash
dir="$(cd -P -- "$(dirname -- "${BASH_SOURCE[0]}")" && pwd -P)"
Note: A double dash (--) is used in commands to signify the end of command options, so files containing dashes or other special characters won't break the command.
Note: In Bash, use ${BASH_SOURCE[0]} in favor of $0, otherwise the path can break when sourcing it (source/.).
*For Linux, Mac and other BSD:
cd "$(dirname "$(realpath -- "$0")")";
Note: realpath should be installed in the most popular Linux distribution by default (like Ubuntu), but in some it can be missing, so you have to install it.
Note: If you're using Bash, use ${BASH_SOURCE[0]} in favor of $0, otherwise the path can break when sourcing it (source/.).
Otherwise you could try something like that (it will use the first existing tool):
cd "$(dirname "$(readlink -f -- "$0" || realpath -- "$0")")"
For Linux specific:
cd "$(dirname "$(readlink -f -- "$0")")"
*Using GNU readlink on BSD/Mac:
cd "$(dirname "$(greadlink -f -- "$0")")"
Note: You need to have coreutils installed
(e.g. 1. Install Homebrew, 2. brew install coreutils).
In bash
In bash you can use Parameter Expansions to achieve that, like:
cd "${0%/*}"
but it doesn't work if the script is run from the same directory.
Alternatively you can define the following function in bash:
realpath () {
[[ "$1" = /* ]] && echo "$1" || echo "$PWD/${1#./}"
}
This function takes 1 argument. If argument has already absolute path, print it as it is, otherwise print $PWD variable + filename argument (without ./ prefix).
or here is the version taken from Debian .bashrc file:
function realpath()
{
f=$#
if [ -d "$f" ]; then
base=""
dir="$f"
else
base="/$(basename -- "$f")"
dir="$(dirname -- "$f")"
fi
dir="$(cd -- "$dir" && /bin/pwd)"
echo "$dir$base"
}
Related:
How to detect the current directory in which I run my shell script?
How do I get the directory where a Bash script is located from within the script itself?
Bash script absolute path with OS X
Reliable way for a Bash script to get the full path to itself
See also:
How can I get the behavior of GNU's readlink -f on a Mac?
cd "$(dirname "${BASH_SOURCE[0]}")"
It's easy. It works.
The accepted answer works well for scripts that have not been symlinked elsewhere, such as into $PATH.
#!/bin/bash
cd "$(dirname "$0")"
However if the script is run via a symlink,
ln -sv ~/project/script.sh ~/bin/;
~/bin/script.sh
This will cd into the ~/bin/ directory and not the ~/project/ directory, which will probably break your script if the purpose of the cd is to include dependencies relative to ~/project/
The symlink safe answer is below:
#!/bin/bash
cd "$(dirname "$(readlink -f "${BASH_SOURCE[0]}")")" # cd current directory
readlink -f is required to resolve the absolute path of the potentially symlinked file.
The quotes are required to support filepaths that could potentially contain whitespace (bad practice, but its not safe to assume this won't be the case)
This script seems to work for me:
#!/bin/bash
mypath=`realpath $0`
cd `dirname $mypath`
pwd
The pwd command line echoes the location of the script as the current working directory no matter where I run it from.
There are a lot of correct answers in here, but one that tends to be more useful for me (making sure a script's relative paths remain predictable/work) is to use pushd/popd:
pushd "$(dirname ${BASH_SOURCE:0})"
trap popd EXIT
# ./xyz, etc...
This will push the source file's directory on to a navigation stack, thereby changing the working directory, but then, when the script exits (for whatever reason, including failure), the trap will run popd, restoring the current working directory before it was executed. If the script were to cd and then fail, your terminal could be left in an unpredictable state after the execution ends - the trap prevents this.
I take this and it works.
#!/bin/bash
cd "$(dirname "$0")"
CUR_DIR=$(pwd)
Get the real path to your script
if [ -L $0 ] ; then
ME=$(readlink $0)
else
ME=$0
fi
DIR=$(dirname $ME)
(This is answer to the same my question here: Get the name of the directory where a script is executed)
cd "`dirname $(readlink -f ${0})`"
Most answers either don't handle files which are symlinked via a relative path, aren't one-liners or don't handle BSD (Mac). A solution which does all three is:
HERE=$(cd "$(dirname "$BASH_SOURCE")"; cd -P "$(dirname "$(readlink "$BASH_SOURCE" || echo .)")"; pwd)
First, cd to bash's conception of the script's directory. Then readlink the file to see if it is a symlink (relative or otherwise), and if so, cd to that directory. If not, cd to the current directory (necessary to keep things a one-liner). Then echo the current directory via pwd.
You could add -- to the arguments of cd and readlink to avoid issues of directories named like options, but I don't bother for most purposes.
You can see the full explanation with illustrations here:
https://www.binaryphile.com/bash/2020/01/12/determining-the-location-of-your-script-in-bash.html
echo $PWD
PWD is an environment variable.
If you just need to print present working directory then you can follow this.
$ vim test
#!/bin/bash
pwd
:wq to save the test file.
Give execute permission:
chmod u+x test
Then execute the script by ./test then you can see the present working directory.