I have larger shell script which handles different things.
It will get it's own location by the following...
BASEDIR=`dirname $0`/..
BASEDIR=`(cd "$BASEDIR"; pwd)`
then BASEDIR will be used create other variables like
REPO="$BASEDIR"/repo
But the problem is that this shell script does not work if the path contains spaces where it is currently executed.
So the question is: Does exist a good solution to solve that problem ?
Be sure to double-quote anything that may contain spaces:
BASEDIR="`dirname $0`"
BASEDIR="`(cd \"$BASEDIR\"; pwd)`"
The answer is "Quotes everywhere."
If the path you pass in has a space in it then dirname $0 will fail.
$ cat quote-test.sh
#!/bin/sh
test_dirname_noquote () {
printf 'no quotes: '
dirname $1
}
test_dirname_quote () {
printf 'quotes: '
dirname "$1"
}
test_dirname_noquote '/path/to/file with spaces/in.it'
test_dirname_quote '/path/to/file with spaces/in.it'
$ sh quote-test.sh
no quotes: usage: dirname string
quotes: /path/to/file with spaces
Also, try this fun example
#!/bin/sh
mkdir -p /tmp/foo/bar/baz
cd /tmp/foo
ln -s bar quux
cd quux
cat >>find-me.sh<<"."
#!/bin/sh
self_dir="$(dirname $0)"
base_dir="$( (cd "$self_dir/.." ; pwd -P) )"
repo="$base_dir/repo"
printf 'self: %s\n' "$self_dir"
printf 'base: %s\n' "$base_dir"
printf 'repo: %s\n' "$repo"
.
sh find-me.sh
rm -rf /tmp/foo
Result when you run it:
$ sh example.sh
self: .
base: /tmp/foo
repo: /tmp/foo/repo
Quote your full variable like this:
REPO="$BASEDIR/repo"
There is no reliable and/or portable way to do this correctly.
See How do I determine the location of my script? as to why
The best answer is the following, which is still OS dependent
BASEDIR=$(readlink -f $0)
Then you can do things like REPO="$BASEDIR"/repo , just be sure to quote your variables as you did.
Works perfectly fine for me. How are you using REPO? What specifically "doesn't work" for you?
I tested
#!/bin/sh
BASEDIR=`dirname $0`/..
BASEDIR=`(cd "$BASEDIR"; pwd)`
REPO="$BASEDIR"/repo
echo $REPO
in a ".../a b/c d" directory. It outputs ".../a b/repo", as expected.
Please give the specific error that you are receiving... A "doesn't work" bug report is the least useful bug report, and every programmer absolutely hates it.
Using spaces in directory names in unix is always an issue so if they can be avoided by using underscores, this prevents lots of strange scripting behaviour.
I'm unclear why you are setting BASEDIR to be the parent directory of the directory containing the current script (..) and then resetting it after changing into that directory
The path to the directory should still work if it has ..
e.g. /home/trevor/data/../repo
BASEDIR=`dirname $0`/..
I think if you echo out $REPO it should have the path correctly assigned because you used quotes when assigning it but if you then try to use $REPO somewhere else in the script, you will need to use double quotes around that too.
e.g.
#!/bin/ksh
BASEDIR=`dirname $0`/..
$REPO="$BASEDIR"/repo
if [ ! -d ["$REPO"]
then
echo "$REPO does not exist!"
fi
Use speech marks as below:
BASEDIR=`dirname "${0}"`/..
Related
My script:
#!/usr/bin/env bash
PATH=/home/user/example/foo/bar
mkdir -p /tmp/backup$PATH
And now I want to get first folder of "$PATH": /home/
cd /tmp/backup
rm -rf ./home/
cd - > /dev/null
How can I always detect the first folder like the example above? "dirname $PATH" just returns "/home/user/example/foo/".
Thanks in advance! :)
I've found a solution:
#/usr/bin/env bash
DIRECTORY="/home/user/example/foo/bar"
BASE_DIRECTORY=$(echo "$DIRECTORY" | cut -d "/" -f2)
echo "#$BASE_DIRECTORY#";
This returns always the first directory. In this example it would return following:
#home#
Thanks to #condorwasabi for his idea with awk! :)
You can try this awk command:
basedirectory=$(echo "$PATH" | awk -F "/" '{print $2}')
At this point basedirectory will be the string home
Then you write:
rm -rf ./"$basedirectory"/
If PATH always has an absolute form you can do tricks like
ROOT=${PATH#/} ROOT=/${ROOT%%/*}
Or
IFS=/ read -ra T <<< "$PATH"
ROOT=/${T[1]}
However I should also add to that that it's better to use other variables and not to use PATH as it would alter your search directories for binary files, unless you really intend to.
Also you can opt to convert your path to absolute form through readlink -f or readlink -m:
ABS=$(readlink -m "$PATH")
You can also refer to my function getabspath.
To get the first directory component of VAR:
echo ${VAR%${VAR#/*/}}
So, if VAR="/path/to/foo", this returns /path/.
Explanation:
${VAR#X} strips off the prefix X and returns the remainder. So if VAR=/path/to/foo, then /*/ matches the prefix /path/ and the expression returns the suffix to/foo.
${VAR%X} strips off the suffix X. By inserting the output of ${VAR#X}, it strips off the suffix and returns the prefix.
If you can guarantee that your paths are well formed this is a convenient method. It won't work well for some paths, such as //path/to/foo or path/to/foo, but you can handle such cases by breaking down the strings further.
To get the first firectory:
path=/home/user/example/foo/bar
mkdir -p "/tmp/backup$path"
cd /tmp/backup
arr=( */ )
echo "${arr[0]}"
PS: Never use PATH variable in your script as it will overrider default PATH and you script won't be able to execute many system utilities
EDIT: Probably this should work for you:
IFS=/ && set -- $path; echo "$2"
home
Pure bash:
DIR="/home/user/example/foo/bar"
[[ "$DIR" =~ ^[/][^/]+ ]] && printf "$BASH_REMATCH"
Easy to tweak the regex.
You can use dirname...
#/usr/bin/env bash
DIRECTORY="/home/user/example/foo/bar"
BASE_DIRECTORY=$(dirname "${DIRECTORY}")
echo "#$BASE_DIRECTORY#";
Outputs the following...
/home/user/example/foo
I'm trying to learn a bit about bash-commands -- more specifically about backup-scripts.
Unfortunately, I get a syntax error on the last line. I have no idea what I've done wrong and would appreciate any feedback on this matter.
Code:
#!/bin/sh
if [ -f $"/archive/backup-20110111.tar.gz" ]; then
echo "File already exists"
else
sudo cp /home/plepple/Documents/backup/backup-20110111.tar.gz
/home/plepple/Documents/backup/archive/backup-20110111.tar.gz
rm /home/plepple/Documents/backup/backup-20110111.tar.gz
fi
if [ -f $"/archive/backup-20110112.tar.gz" ]; then
echo "File already exists"
else
sudo cp /home/plepple/Documents/backup/backup-20110112.tar.gz
/home/plepple/Documents/backup/archive/backup-20110112.tar.gz
rm /home/plepple/Documents/backup/backup-20110112.tar.gz
fi
curdate='date +%Y%m%d'
mv /home/plepple/Documents/backup/backup.tar.gz
/home/plepple/Documents/backup/backup-$curdate.tar.gz
I tried to execute it (through bash) with:
bash backupscript.sh
All the files and directories exist.
Thanks!
mv /home/plepple/Documents/backup/backup.tar.gz
/home/plepple/Documents/backup/backup-$datum.tar.gz
should be
mv /home/plepple/Documents/backup/backup.tar.gz \
/home/plepple/Documents/backup/backup-$datum.tar.gz
The same goes for
sudo cp /home/plepple/Documents/backup/backup-20110111.tar.gz \
/home/plepple/Documents/backup/archive/backup-20110111.tar.gz
and
sudo cp /home/plepple/Documents/backup/backup-20110112.tar.gz \
/home/plepple/Documents/backup/archive/backup-20110112.tar.gz
Where are the fis to end the if blocks?
This isn't the problem, just a misc bash syntax correction: in bash, the construct $"somestring" invokes localization. From the bash manpage:
A double-quoted string preceded by a dollar sign ($) will cause the
string to be translated according to the current locale. If the cur-
rent locale is C or POSIX, the dollar sign is ignored. If the string
is translated and replaced, the replacement is double-quoted.
That doesn't appear to be relevant to the filepaths in your if tests, so you should probably leave the $ off. Actually, since the filepaths don't have any funny characters in them, you don't even need the quotes around them (although overuse of double-quotes is much better than underuse).
#!/bin/sh
files = 'ls /myDir/myDir2/myDir3/'
for file in $files do
echo $file
java myProg $file /another/directory/
done
What i'm trying to do is iterate through every file name under /myDir/myDir2/myDir3/, then use that file name as the first argument in calling a java program (second argument is "/another/directory")
When I run this script: . myScript.sh
I get this error:
-bash: files: command not found
What did I do wrong in my script? Thanks!
Per Neeaj's answer, strip off the whitespace from files =.
Better yet, use:
#!/bin/sh -f
dir=/myDir/MyDir2/MyDir3
for path in $dir/*; do
file=$(basename $path)
echo "$file"
java myProg "$file" arg2 arg3
done
Bash is perfectly capable of expanding the * wildcard itself, without spawning a copy of ls to do the job for it!
EDIT: changed to call basename rather than echo to meet OP's (previously unstated) requirement that the path echoed be relative and not absolute. If the cwd doesn't matter, then even better I'd go for:
#!/bin/sh -f
cd /myDir/MyDir2/MyDir3
for file in *; do
echo "$file"
java myProg "$file" arg2 arg3
done
and avoid the calls to basename altogether.
strip off the whitespace in and after files = as files=RHS of assignment
Remove the space surrounding the '=' : change
files = 'ls /myDir/myDir2/myDir3/'
into:
files='ls /myDir/myDir2/myDir3/'
and move the 'do' statement to its own line:
for file in $files
do
....
quote your variables and no need to use ls.
#!/bin/sh
for file in /myDir/myDir2/*
do
java myProg "$file" /another/directory/
done
How could I retrieve the current working directory/folder name in a bash script, or even better, just a terminal command.
pwd gives the full path of the current working directory, e.g. /opt/local/bin but I only want bin.
No need for basename, and especially no need for a subshell running pwd (which adds an extra, and expensive, fork operation); the shell can do this internally using parameter expansion:
result=${PWD##*/} # to assign to a variable
result=${result:-/} # to correct for the case where PWD=/
printf '%s\n' "${PWD##*/}" # to print to stdout
# ...more robust than echo for unusual names
# (consider a directory named -e or -n)
printf '%q\n' "${PWD##*/}" # to print to stdout, quoted for use as shell input
# ...useful to make hidden characters readable.
Note that if you're applying this technique in other circumstances (not PWD, but some other variable holding a directory name), you might need to trim any trailing slashes. The below uses bash's extglob support to work even with multiple trailing slashes:
dirname=/path/to/somewhere//
shopt -s extglob # enable +(...) glob syntax
result=${dirname%%+(/)} # trim however many trailing slashes exist
result=${result##*/} # remove everything before the last / that still remains
result=${result:-/} # correct for dirname=/ case
printf '%s\n' "$result"
Alternatively, without extglob:
dirname="/path/to/somewhere//"
result="${dirname%"${dirname##*[!/]}"}" # extglob-free multi-trailing-/ trim
result="${result##*/}" # remove everything before the last /
result=${result:-/} # correct for dirname=/ case
Use the basename program. For your case:
% basename "$PWD"
bin
$ echo "${PWD##*/}"
Use:
basename "$PWD"
OR
IFS=/
var=($PWD)
echo ${var[-1]}
Turn the Internal Filename Separator (IFS) back to space.
IFS=
There is one space after the IFS.
You can use a combination of pwd and basename. E.g.
#!/bin/bash
CURRENT=`pwd`
BASENAME=`basename "$CURRENT"`
echo "$BASENAME"
exit;
How about grep:
pwd | grep -o '[^/]*$'
This thread is great! Here is one more flavor:
pwd | awk -F / '{print $NF}'
basename $(pwd)
or
echo "$(basename $(pwd))"
I like the selected answer (Charles Duffy), but be careful if you are in a symlinked dir and you want the name of the target dir. Unfortunately I don't think it can be done in a single parameter expansion expression, perhaps I'm mistaken. This should work:
target_PWD=$(readlink -f .)
echo ${target_PWD##*/}
To see this, an experiment:
cd foo
ln -s . bar
echo ${PWD##*/}
reports "bar"
DIRNAME
To show the leading directories of a path (without incurring a fork-exec of /usr/bin/dirname):
echo ${target_PWD%/*}
This will e.g. transform foo/bar/baz -> foo/bar
echo "$PWD" | sed 's!.*/!!'
If you are using Bourne shell or ${PWD##*/} is not available.
Surprisingly, no one mentioned this alternative that uses only built-in bash commands:
i="$IFS";IFS='/';set -f;p=($PWD);set +f;IFS="$i";echo "${p[-1]}"
As an added bonus you can easily obtain the name of the parent directory with:
[ "${#p[#]}" -gt 1 ] && echo "${p[-2]}"
These will work on Bash 4.3-alpha or newer.
There are a lots way of doing that I particularly liked Charles way because it avoid a new process, but before know this I solved it with awk
pwd | awk -F/ '{print $NF}'
For the find jockeys out there like me:
find $PWD -maxdepth 0 -printf "%f\n"
i usually use this in sh scripts
SCRIPTSRC=`readlink -f "$0" || echo "$0"`
RUN_PATH=`dirname "${SCRIPTSRC}" || echo .`
echo "Running from ${RUN_PATH}"
...
cd ${RUN_PATH}/subfolder
you can use this to automate things ...
Just use:
pwd | xargs basename
or
basename "`pwd`"
Below grep with regex is also working,
>pwd | grep -o "\w*-*$"
If you want to see only the current directory in the bash prompt region, you can edit .bashrc file in ~. Change \w to \W in the line:
PS1='${debian_chroot:+($debian_chroot)}\[\033[01;32m\]\u#\h\[\033[00m\]:\[\033[01;34m\]\w\[\033[00m\]\$ '
Run source ~/.bashrc and it will only display the directory name in the prompt region.
Ref: https://superuser.com/questions/60555/show-only-current-directory-name-not-full-path-on-bash-prompt
I strongly prefer using gbasename, which is part of GNU coreutils.
Just run the following command line:
basename $(pwd)
If you want to copy that name:
basename $(pwd) | xclip -selection clipboard
An alternative to basname examples
pwd | grep -o "[^/]*$"
OR
pwd | ack -o "[^/]+$"
My shell did not come with the basename package and I tend to avoid downloading packages if there are ways around it.
You can use the basename utility which deletes any prefix ending in / and the suffix (if present in string) from string, and prints the
result on the standard output.
$basename <path-of-directory>
Just remove any character until a / (or \, if you're on Windows). As the match is gonna be made greedy it will remove everything until the last /:
pwd | sed 's/.*\///g'
In your case the result is as expected:
λ a='/opt/local/bin'
λ echo $a | sed 's/.*\///g'
bin
Here's a simple alias for it:
alias name='basename $( pwd )'
After putting that in your ~/.zshrc or ~/.bashrc file and sourcing it (ex: source ~/.zshrc), then you can simply run name to print out the current directories name.
The following commands will result in printing your current working directory in a bash script.
pushd .
CURRENT_DIR="`cd $1; pwd`"
popd
echo $CURRENT_DIR
Is it possible to find out the full path to the script that is currently executing in KornShell (ksh)?
i.e. if my script is in /opt/scripts/myscript.ksh, can I programmatically inside that script discover /opt/scripts/myscript.ksh ?
Thanks,
You could use:
## __SCRIPTNAME - name of the script without the path
##
typeset -r __SCRIPTNAME="${0##*/}"
## __SCRIPTDIR - path of the script (as entered by the user!)
##
__SCRIPTDIR="${0%/*}"
## __REAL_SCRIPTDIR - path of the script (real path, maybe a link)
##
__REAL_SCRIPTDIR=$( cd -P -- "$(dirname -- "$(command -v -- "$0")")" && pwd -P )
In korn shell, all of these $0 solutions fail if you are sourcing in the script in question. The correct way to get what you want is to use $_
$ cat bar
echo dollar under is $_
echo dollar zero is $0
$ ./bar
dollar under is ./bar
dollar zero is ./bar
$ . ./bar
dollar under is bar
dollar zero is -ksh
Notice the last line there? Use $_. At least in Korn. YMMV in bash, csh, et al..
Well it took me a while but this one is so simple it screams.
_SCRIPTDIR=$(cd $(dirname $0);echo $PWD)
since the CD operates in the spawned shell with $() it doesn't affect the current script.
How the script was called is stored in the variable $0. You can use readlink to get the absolute file name:
readlink -f "$0"
The variable $RPATH contains the relative path to the real file or the real path for a real file.
CURPATH=$( cd -P -- "$(dirname -- "$(command -v -- "$0")")" && pwd -P )
CURLOC=$CURPATH/`basename $0`
if [ `ls -dl $CURLOC |grep -c "^l" 2>/dev/null` -ne 0 ];then
ROFFSET=`ls -ld $CURLOC|cut -d ">" -f2 2>/dev/null`
RPATH=`ls -ld $CURLOC/$ROFFSET 2>/dev/null`
else
RPATH=$CURLOC
fi
echo $RPATH
This is what I did:
if [[ $0 != "/"* ]]; then
DIR=`pwd`/`dirname $0`
else
DIR=`dirname $0`
fi
readlink -f would be the best if it was portable, because it resolves every links found for both directories and files.
On mac os x there is no readlink -f (except maybe via macports), so you can only use readlink to get the destination of a specific symbolic link file.
The $(cd -P ... pwd -P) technique is nice but only works to resolve links for directories leading to the script, it doesn't work if the script itself is a symlink
Also, one case that wasn't mentioned : when you launch a script by passing it as an argument to a shell (/bin/sh /path/to/myscript.sh), $0 is not usable in this case
I took a look to mysql "binaries", many of them are actually shell scripts ; and now i understand why they ask for a --basedir option or need to be launched from a specific working directory ; this is because there is no good solution to locate the targeted script
This works also, although it won't give the "true" path if it's a link. It's simpler, but less exact.
SCRIPT_PATH="$(whence ${0})"
Try which command.
which scriptname
will give you the full qualified name of the script along with its absolute path
I upgraded the Edward Staudt's answer, to be able to deal with absolute-path symbolic links, and with chains of links too.
DZERO=$0
while true; do
echo "Trying to find real dir for script $DZERO"
CPATH=$( cd -P -- "$(dirname -- "$(command -v -- "$DZERO")")" && pwd -P )
CFILE=$CPATH/`basename $DZERO`
if [ `ls -dl $CFILE | grep -c "^l" 2>/dev/null` -eq 0 ];then
break
fi
LNKTO=`ls -ld $CFILE | cut -d ">" -f2 | tr -d " " 2>/dev/null`
DZERO=`cd $CPATH ; command -v $LNKTO`
done
Ugly, but works...
After run this, the path is $CPATH and the file is $CFILE
Try using this:
dir = $(dirname $0)
Using $_ provides the last command.
>source my_script
Works if I issue the command twice:
>source my_script
>source my_script
If I use a different sequence of commands:
>who
>source my_script
The $_ variable returns "who"