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 ${workdir} work
The command ln -s is the command, creating the symlink.
If you don't like the creation of that symlink, you might put that line in comment (don't delete it: in case you're not satisfied, it's easier to uncomment it).
You can solve your issue, using this command:
ln -sTf ...
This removes the existing destination files beforehand.
Related
I want to translate this bash-script intro a zsh-script. Hence I have no experience with this I hope I may get help here:
bash script:
SCRIPT_PATH="${BASH_SOURCE[0]}";
if([ -h "${SCRIPT_PATH}" ]) then
while([ -h "${SCRIPT_PATH}" ]) do SCRIPT_PATH=`readlink "${SCRIPT_PATH}"`; done
fi
pushd . > /dev/null
cd `dirname ${SCRIPT_PATH}` > /dev/null
SCRIPT_PATH=`pwd`;
popd > /dev/null
What I already know is that I can use
SCRIPT_PATH="$0"; to get the path were the script is located at. But then I get errors with the "readlink" statement.
Thanks for your help
Except for BASH_SOURCE I see no changes that you need to make. But what is the purpose of the script? If you want to get directory your script is located at there is ${0:A:h} (:A will resolve all symlinks, :h will truncate last path component leaving you with a directory name):
SCRIPT_PATH="${0:A:h}"
and that’s all. Note that original script has something strange going on:
if(…) and while(…) launch … in a subshell. You do not need subshell here, it is faster to do these checks using just if … and while ….
pushd . is not needed at all. While using pushd you normally replace the cd call with it:
pushd "$(dirname $SCRIPT_PATH)" >/dev/null
SCRIPT_PATH="$(pwd)"
popd >/dev/null
cd `…` will fail if … outputs something with spaces. It is possible for a directory to contain a space. In the above example I use "$(…)", "`…`" will also work.
You do not need trailing ; in variable declarations.
There is readlink -f that will resolve all symlinks thus you may consider reducing original script to SCRIPT_PATH="$(dirname $(readlink -f "${BASH_SOURCE[0]}"))" (the behavior may change as your script seems to resolve symlinks only in last component): this is bash equivalent to ${0:A:h}.
if [ -h "$SCRIPT_PATH" ] is redundant since while body with the same condition will not be executed unless script path is a symlink.
readlink $SCRIPT_PATH will return symlink relative to the directory containing $SCRIPT_PATH. Thus original script cannot possibly used to resolve symlinks in last component.
There is no ; between if(…) and then. I am surprised bash accepts this.
All of the above statements apply both to bash and zsh.
If resolving only symlinks only in last component is essential you should write it like this:
SCRIPT_PATH="$0:a"
function ResolveLastComponent()
{
pushd "$1:h" >/dev/null
local R="$(readlink "$1")"
R="$R:a"
popd >/dev/null
echo $R
}
while test -h "$SCRIPT_PATH" ; do
SCRIPT_PATH="$(ResolveLastComponent "$SCRIPT_PATH")"
done
.
To illustrate 7th statement there is the following example:
Create directory $R/bash ($R is any directory, e.g. /tmp).
Put your script there without modifications, e.g. under name $R/bash/script_path.bash. Add line echo "$SCRIPT_PATH" at the end of it and line #!/bin/bash at the start for testing.
Make it executable: chmod +x $R/bash/script_path.bash.
Create a symlink to it: cd $R/bash && ln -s script_path.bash link.
cd $R
Launch $R/bash/1. Now you will see that your script outputs $R while it should output $R/bash like it does when you launch $R/bash/script_path.bash.
I wrote this little shell script(test.sh) to create a basic folder structure:
#!/bin/bash
# Check if directory already exists,
# if it doesnt, create one.
if [ ! -d "~/.dir1" ]; then
mkdir ".dir1"
else
rm -rf ".dir1"
mkdir ".dir1"
fi
When I run
test.sh
in console, the hidden folder is created.
But:
When I run it again it tells me:
mkdir: .dir1: File exists
But it could exist because I removed it in my shell script before I created a new one!
So why does it display this message?
Thanks and greetings!
Replace
[ ! -d "~/.dir1" ]
by
[ ! -d "${HOME}/.dir1" ]
I would simply use -p.
mkdir -p "$HOME/dir1"
If you pass -p, mkdir wouldn't throw an error if the directory already exists, it would simply silently return in that case.
If you want to make sure folder is empty use this:
rm -rf "$HOME/dir1"
mkdir -p "$HOME/dir1"
and no if! The basic problem with the if is the fact that it is not immune against race conditions. When the script went off from CPU right after the if - and creates "dir1" - your script will fail when it enters the CPU again since it still thinks the directory does not exist.
What you are doing by "~/.dir1" is not right. It's just another string for a directory name literally "~/.dir1" i.e ~ is not being expanded to $HOME.
Use full path or ~/".dir1" or ~/.dir1 instead.
You can use $HOME too: $HOME/.dir1 or "$HOME/.dir1" or "$HOME"/".dir1" all of them will produce same result... but quoting variables is a good practice.
~ isn't expanded when you place it in quotes. You need to leave it unquoted.
if [ ! -d ~/.dir1 ]
Of note, you're checking for ~/.dir1 but you make .dir1. That's only the same directory if the current directory is ~. If it isn't, they're not the same.
Also, mkdir -p will do this for you, creating a directory only if it doesn't exist already. You could simplify your script to:
mkdir -p ~/.dir1
or
rm -rf ~/.dir1
mkdir ~/.dir1
I want to translate this bash-script intro a zsh-script. Hence I have no experience with this I hope I may get help here:
bash script:
SCRIPT_PATH="${BASH_SOURCE[0]}";
if([ -h "${SCRIPT_PATH}" ]) then
while([ -h "${SCRIPT_PATH}" ]) do SCRIPT_PATH=`readlink "${SCRIPT_PATH}"`; done
fi
pushd . > /dev/null
cd `dirname ${SCRIPT_PATH}` > /dev/null
SCRIPT_PATH=`pwd`;
popd > /dev/null
What I already know is that I can use
SCRIPT_PATH="$0"; to get the path were the script is located at. But then I get errors with the "readlink" statement.
Thanks for your help
Except for BASH_SOURCE I see no changes that you need to make. But what is the purpose of the script? If you want to get directory your script is located at there is ${0:A:h} (:A will resolve all symlinks, :h will truncate last path component leaving you with a directory name):
SCRIPT_PATH="${0:A:h}"
and that’s all. Note that original script has something strange going on:
if(…) and while(…) launch … in a subshell. You do not need subshell here, it is faster to do these checks using just if … and while ….
pushd . is not needed at all. While using pushd you normally replace the cd call with it:
pushd "$(dirname $SCRIPT_PATH)" >/dev/null
SCRIPT_PATH="$(pwd)"
popd >/dev/null
cd `…` will fail if … outputs something with spaces. It is possible for a directory to contain a space. In the above example I use "$(…)", "`…`" will also work.
You do not need trailing ; in variable declarations.
There is readlink -f that will resolve all symlinks thus you may consider reducing original script to SCRIPT_PATH="$(dirname $(readlink -f "${BASH_SOURCE[0]}"))" (the behavior may change as your script seems to resolve symlinks only in last component): this is bash equivalent to ${0:A:h}.
if [ -h "$SCRIPT_PATH" ] is redundant since while body with the same condition will not be executed unless script path is a symlink.
readlink $SCRIPT_PATH will return symlink relative to the directory containing $SCRIPT_PATH. Thus original script cannot possibly used to resolve symlinks in last component.
There is no ; between if(…) and then. I am surprised bash accepts this.
All of the above statements apply both to bash and zsh.
If resolving only symlinks only in last component is essential you should write it like this:
SCRIPT_PATH="$0:a"
function ResolveLastComponent()
{
pushd "$1:h" >/dev/null
local R="$(readlink "$1")"
R="$R:a"
popd >/dev/null
echo $R
}
while test -h "$SCRIPT_PATH" ; do
SCRIPT_PATH="$(ResolveLastComponent "$SCRIPT_PATH")"
done
.
To illustrate 7th statement there is the following example:
Create directory $R/bash ($R is any directory, e.g. /tmp).
Put your script there without modifications, e.g. under name $R/bash/script_path.bash. Add line echo "$SCRIPT_PATH" at the end of it and line #!/bin/bash at the start for testing.
Make it executable: chmod +x $R/bash/script_path.bash.
Create a symlink to it: cd $R/bash && ln -s script_path.bash link.
cd $R
Launch $R/bash/1. Now you will see that your script outputs $R while it should output $R/bash like it does when you launch $R/bash/script_path.bash.
I need to write a script that will recreate my opt folder if it gets deleted when I remove a package from it. Here's a link to my previous post: dpkg remove to stop processes
Now, the issue I'm running into could be better described here: http://lists.debian.org/debian-devel/2006/03/msg00242.html
I was thinking of just adding a postrem script which checks if an opt directory exists, and if not, creates one. My experience with shell scripts is pretty limited though..
[ -d "$dir" ] || mkdir -p "$dir"
This could be written more verbosely / clearly as:
if ! test -d "$dir"; then
mkdir -p "$dir"
fi
See help test for more information.
I'm trying to store all my profile configuration files (~/.xxx) in git. I'm pretty horrible at bash scripting but I imagine this will be pretty straight forward for you scripting gurus.
Basically, I'd like a script that will create symbolic links in my home directory to files in my repo. Twist is, I'd like it warn and prompt for overwrite if the symlink will be overwriting an actual file. It should also prompt if a sym link is going to be overwritten, but the target path is different.
I don't mind manually editing the script for each link I want to create. I'm more concerned with being able to quickly deploy new config scripts by running this script stored in my repo.
Any ideas?
The ln command is already conservative about erasing, so maybe the KISS approach is good enough for you:
ln -s git-stuff/home/.[!.]* .
If a file or link already exists, you'll get an error message and this link will be skipped.
If you want the files to have a different name in your repository, pass the -n option to ln so that it doesn't accidentally create a symlink in an existing subdirectory of that name:
ln -sn git-stuff/home/profile .profile
...
If you also want to have links in subdirectories of your home directory, cp -as reproduces the directory structure but creates symbolic links for regular files. With the -i option, it prompts if a target already exists.
cp -i -as git-stuff/home/.[!.]* .
(My answer assumes GNU ln and GNU cp, such as you'd find on Linux (and Cygwin) but usually not on other unices.)
The following has race conditions, but it is probably as safe as you can get without filesystem transactions:
# create a symlink at $dest pointing to $source
# not well tested
set -e # abort on errors
if [[ ( -h $dest && $(readlink -n "$dest") != $source ) || -f $dest || -d $dest ]]
then
read -p "Overwrite $dest? " answer
else
answer=y
fi
[[ $answer == y ]] && ln -s -n -f -v -- "$source" "$dest"