I wrote a simple bash script (in /homedir) to run an executable and then move the outputs to /workdir. I also made a soft link of /workdir named work to /homedir for me to switch easily between folders.
All steps are working well, except that an unspecified soft link named 'grids' is created in /workdir to itself. I can't delete it otherwise all outputs are gone as well.
How can this happen?
#!/bin/bash
cd ..
expname=`basename "$PWD"`
echo 'experiment name: '$expname
homedir=/home/b/b380963/icon_foehn/$expname/grids/
workdir=/work/bb1096/b380963/icon_foehn/$expname/grids/
if [ ! -d ${workdir} ]; then
mkdir -p ${workdir}
fi
cd $homedir
ln -s ${workdir} work
cd /home/b/b380963/nwp/dwd_icon_tools_v2/icontools/
./icongridgen --nml $homedir/gridgen_MCH_july.nml
mv ICON_1E_* $workdir/
mv base_grid* $workdir/
It's quite easy to see in your code:
workdir=/work/bb1096/b380963/icon_foehn/$expname/grids/
...
ln -s ${workdir} work
The command ln -s is the command, creating the symlink.
If you don't like the creation of that symlink, you might put that line in comment (don't delete it: in case you're not satisfied, it's easier to uncomment it).
You can solve your issue, using this command:
ln -sTf ...
This removes the existing destination files beforehand.
I have installed Anaconda a few months ago but then uninstalled it and removed all anaconda files by using
rm -rf ~/anaconda
but when I run
echo $PATH
it still outputs a path that point to an Anaconda folder but when I search for it, it doesn't even exist, why is that happening?
What makes you think that non-existent directory are automatically
removed from $PATH? They are not. As an example I can make a new dir
and go there:
$ mkdir /tmp/new-path-dir && cd /tmp/new-path-dir
Add it to the $PATH:
$ PATH=/tmp/new-path-dir:$PATH
$ echo $PATH
/tmp/new-path-dir:<REST_OF_PATH>
Make a new olleh.so (hello spelled backwards) executable inside
it:
$ echo 'echo hi' > olleh.so && chmod +x olleh.so
Then go back to ~:
$ cd ~
And start a olleh.so:
$ olleh.so
hi
Now I can safely remove /tmp/new-path-dir:
$ rm -r /tmp/new-path-dir/
And it still will be shown in my $PATH:
$ echo $PATH
/tmp/new-path-dir:<REST_OF_PATH>
But I won't be able to run olleh.so any more:
$ olleh.so
bash: /tmp/new-path-dir/olleh.so: No such file or directory
And as paths to executables are cached by bash I can get rid of
olleh.so permanently like this:
$ hash -r
$ olleh.so
bash: olleh.so: command not found
As many others have done, I want to create a repo to store my dotfile customizations. Instead of doing ln -s manually, I am using the following script to set things up.
#!/bin/bash
set -e
DIR="$HOME/Documents/Dotfiles"
OLDDIR="$HOME/Documents/Other\ Files/Dotfiles_old"
FILES=($HOME/.bash_profile)
echo "Creating $OLDDIR for backup of any existing dotfiles in ~"
mkdir -p "$OLDDIR"
echo "…done"
echo "Changing to the $DIR directory"
cd "$DIR"
echo "…done"
for FILE in "${FILES[#]}"; do
echo "Backup dotfile $FILE from ~/ to $OLDDIR"
cp -L "$HOME/$FILE" "$OLDDIR"
done
for FILE in "${FILES[#]}"; do
echo "copy $FILE from ~ to $DIR."
cp -L "$HOME/$FILE $DIR/"
echo "Creating symlink to $FILE from ~ to $DIR."
ln -sfn "$DIR/$FILE" "$HOME/$FILE";
done
shellcheck source "$HOME/.bash_profile"
When I run this, cp fails because it thinks that .bash_profile isn't there, which obviously isn't the case:
I think my path to the files may be incorrect, although shellcheck reports nothing. What am I forgetting here?
UPDATE: Made another run at this - minus the cp. The one thing I am still unsure of is the use of exit, in particular since I'm already using -e to check for errors.
Shellcheck and bash -n return 0.
#!/bin/bash
set -e
function makeFiles() {
touch .bash_profile \
touch .gitconfig \
touch .gitignore_global
}
function makeLinks() {
ln -sfn ~/Documents/Dotfiles/.bash_profile ~/.bash_profile \
ln -sfn ~/Documents/Dotfiles/.gitconfig ~/.gitconfig \
ln -sfn ~/Documents/Dotfiles/.gitignore_global ~/.gitignore_global \
source ~/.bash_profile
}
read -rp "This may overwrite existing files. Are you sure? (y/n) " -n 1;
echo "";
if [[ $REPLY =~ ^[Yy]$ ]]; then
makeFiles && makeLinks
fi;
Sigh, ln decides that .bash_profile needs to be a directory for some crazy reason.
You're building the path of the dotfile incorrectly - $FILE already contains the full path of the dotfile, and there's no need to prepend $HOME again. Try with this cp command:
cp -L "$FILE $DIR/"
I have this in my makefile,
rcFiles = .vim .vimrc .gitconfig .hgrc .screenrc .Xresources .dircolors .bashrc .ctags .bash_completion.d
install:
#$(foreach f,$(rcFiles), [ -f $(HOME)/$f ] || ln -v -s $(PWD)/$f $(HOME)/ ; )
if .bashrc exits and I try
make install
I get
ln: creating symbolic link `/home/user/.vim': File exists
ln: creating symbolic link `/home/user/.bash_completion.d': File exists
and the process is aborted.
why no prevented this problem the conditional?
ln -sfvn source target
The --force flag makes it replace an existing link
The --no-dereference avoids creating 'subdirectory' links for links to directory, if the link existed already (useful for the .bash_completion.d and .vim dirs)
rcFiles = .vim .vimrc .gitconfig .hgrc .screenrc .Xresources .dircolors .bashrc .ctags .bash_completion.d
install:
#$(foreach f,$(rcFiles), [ -f $(HOME)/$f ] || ln -v -f -n -s $(PWD)/$f $(HOME)/ ; )
Alternatively
#$(foreach f,$(rcFiles), [ -e $(HOME)/$f ] || ln -v -f -n -s $(PWD)/$f $(HOME)/ ; )
To not only detect files (-f) but also directories. You might want to explicitely check for files and directories [ -f ... || -d ... ].
[ -f $(HOME)/$f ]
is true only if $(HOME)/$f is (expands to) a file. The things you're getting errors on (.vim and .bash_completion.d) are directories. Try this instead:
[ -e "$(HOME)/$f" ]
(The double quotes are not strictly necessary, but will save you grief in the event that $(HOME)/$f were to expand to something with shell metacharacters in it.)
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.