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 ${workdir} work
The command ln -s is the command, creating the symlink.
If you don't like the creation of that symlink, you might put that line in comment (don't delete it: in case you're not satisfied, it's easier to uncomment it).
You can solve your issue, using this command:
ln -sTf ...
This removes the existing destination files beforehand.
I want to translate this bash-script intro a zsh-script. Hence I have no experience with this I hope I may get help here:
bash script:
SCRIPT_PATH="${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.
There are two shell commands:
[[ ! -d $TRACEDIR/pattrace/rpt/.tracking ]] && mkdir $TRACEDIR/pattrace/rpt/.tracking
[[ ! -d $TRACEDIR/pattrace/rpt/.tracking ]] && mkdir -p $TRACEDIR/pattrace/rpt/.tracking
Obviously, the only difference between those commands is -p flag. But what this flag does in this context?
Thanks.
From the mkdir man page:
-p, --parents
no error if existing, make parent directories as needed
In other words, if the directories needed don't exist, they will be created as required. If the directories already exist, it won't cause an error.
This is a good place to look for man pages (in addition to using google of course)
mkdir with the -p option will create all necessary parent directories of the specified path, should they not exist (see man pages). Also, with -p you won't get an error if the directory itself already exists.
In your particular case, the first command might fail because the test for the complete path is not sufficient. The test will also fail if only, say, $TRACEDIR/ exists but the subsequent mkdir will the fail because it would require $TRACEDIR/pattrace/rpt/ to exist.
The second command will work, because mkdir -p creates all missing directories "in between" as well.
I am writing a shell script to run under the KornShell (ksh) on AIX. I would like to use the mkdir command to create a directory. But the directory may already exist, in which case I do not want to do anything. So I want to either test to see that the directory does not exist, or suppress the "File exists" error that mkdir throws when it tries to create an existing directory.
How can I best do this?
Try mkdir -p:
mkdir -p foo
Note that this will also create any intermediate directories that don't exist; for instance,
mkdir -p foo/bar/baz
will create directories foo, foo/bar, and foo/bar/baz if they don't exist.
Some implementation like GNU mkdir include mkdir --parents as a more readable alias, but this is not specified in POSIX/Single Unix Specification and not available on many common platforms like macOS, various BSDs, and various commercial Unixes, so it should be avoided.
If you want an error when parent directories don't exist, and want to create the directory if it doesn't exist, then you can test for the existence of the directory first:
[ -d foo ] || mkdir foo
This should work:
$ mkdir -p dir
or:
if [[ ! -e $dir ]]; then
mkdir $dir
elif [[ ! -d $dir ]]; then
echo "$dir already exists but is not a directory" 1>&2
fi
which will create the directory if it doesn't exist, but warn you if the name of the directory you're trying to create is already in use by something other than a directory.
Use the -p flag.
man mkdir
mkdir -p foo
Defining complex directory trees with one command
mkdir -p project/{lib/ext,bin,src,doc/{html,info,pdf},demo/stat/a}
If you don't want to show any error message:
[ -d newdir ] || mkdir newdir
If you want to show your own error message:
[ -d newdir ] && echo "Directory Exists" || mkdir newdir
mkdir foo works even if the directory exists.
To make it work only if the directory named "foo" does not exist, try using the -p flag.
Example:
mkdir -p foo
This will create the directory named "foo" only if it does not exist. :)
The old tried and true
mkdir /tmp/qq >/dev/null 2>&1
will do what you want with none of the race conditions many of the other solutions have.
Sometimes the simplest (and ugliest) solutions are the best.
Simple, silent and deadly:
mkdir -p /my/new/dir >/dev/null 2>&1
You can either use an if statement to check if the directory exists or not. If it does not exits, then create the directory.
dir=/home/dir_name
if [ ! -d $dir ]
then
mkdir $dir
else
echo "Directory exists"
fi
You can directory use mkdir with -p option to create a directory. It will check if the directory is not available it will.
mkdir -p $dir
mkdir -p also allows to create the tree structure of the directory. If you want to create the parent and child directories using same command, can opt mkdir -p
mkdir -p /home/parent_dir /home/parent_dir/child1 /home/parent_dir/child2
mkdir does not support -p switch anymore on Windows 8+ systems.
You can use this:
IF NOT EXIST dir_name MKDIR dir_name
directory_name = "foo"
if [ -d $directory_name ]
then
echo "Directory already exists"
else
mkdir $directory_name
fi
This is a simple function (Bash shell) which lets you create a directory if it doesn't exist.
#------------------------------------------#
# Create a directory if it does not exist. #
#------------------------------------------#
# Note the "-p" option in the mkdir #
# command which creates directories #
# recursively. #
#------------------------------------------#
createDirectory() {
mkdir -p -- "$1"
}
You can call the above function as:
createDirectory "$(mktemp -d dir-example.XXXXX)/fooDir/BarDir"
The above creates fooDir and BarDir if they don't exist. Note the "-p" option in the mkdir command which creates directories recursively.
Referring to man page man mkdir for option -p
-p, --parents
no error if existing, make parent directories as needed
which will create all directories in a given path, if exists throws no error otherwise it creates all directories from left to right in the given path. Try the below command. the directories newdir and anotherdir doesn't exists before issuing this command
Correct Usage
mkdir -p /tmp/newdir/anotherdir
After executing the command you can see newdir and anotherdir created under /tmp. You can issue this command as many times you want, the command always have exit(0). Due to this reason most people use this command in shell scripts before using those actual paths.
Or if you want to check for existence first:
if [[ ! -e /path/to/newdir ]]; then
mkdir /path/to/newdir
fi
-e is the exist test for KornShell.
You can also try googling a KornShell manual.
Improvement on the 'classic' solution (by Brian Campbell) - to handle the case of symlink to a directory.
[ -d foo/. ] || mkdir foo
mkdir -p sam
mkdir = Make Directory
-p = --parents
(no error if existing, make parent directories as needed)
if [ !-d $dirName ];then
if ! mkdir $dirName; then # Shorter version. Shell will complain if you put braces here though
echo "Can't make dir: $dirName"
fi
fi