I'm inexperienced with bash scripts, mostly cobbling things together from google searches. The following simple script creates file foo and changes its permission bits to whatever the script's $1 is. Can an experienced bash scripter please advise what would be a proper and/or efficient way to ensure that $1 is valid permission bits, i.e. a valid arg1 to chmod?
#!/bin/sh
#mkfoo.sh
if [ $# -eq 0 ]
then
return
fi
perm=$1
touch ./foo
chmod "${perm}" ./foo
As part of my inexperience, I'm unclear when/why variables are strings or integers. A string-validation approach feels infeasible because of the number of values the permission bits could be. What comes to mind is that I'd want to validate whether $1 is an integer, and if so, that it's between 0 and 777. But I'm unclear if that's a sufficient/efficient validation: I've got a rudimentary understanding of linux file permissions, and I'm led to believe there are lots of subtleties. Grateful for help.
If you only want to allow numeric permissions, you can check them with a pattern check:
case "$perms" in
[0-7][0-7][0-7])
touch ./foo
chmod "${perm}" ./foo
;;
*)
echo "Invalid permissions $perms"
;;
esac
From your comments, your goal is to give up on the chmod if the permissions specified are invalid.
chmod "$1" ./foo 2>/dev/null
2 is the file descriptor for stderr. Redirecting to /dev/null will let it fail silently. chmod, as stated in the comments, does its own validation on if the permissions are acceptable.
This script will accept any valid permissions that chmod allows.
It will create the file (if it doesn't already exist) and attempt to set the permissions. If setting the permissions fails the file is removed.
It requires exactly two arguments; the filename to create and the permissions to set.
This allows you to use symbolic or 4-digit permissions, such as create foo u+x,g+rwx (assuming the script is named "create") or create foo 2640
This is pretty simple, as an example. I often include a usage() function which I would call in place of the first echo. You could also include default permissions if the seconds argument was omitted.
#!/bin/sh
if [ $# -ne 2 ]
then
echo "output a usage message, don't just return"
exit
fi
if [ -e "${1}" ]
then
echo "${1} already exists"
exit
fi
touch ./"${1}"
if chmod "${2}" "${1}" > /dev/null 2>&1
then
echo "${1} created with permissions ${2}"
else
rm "${1}"
echo "${1} not created: invalid permissions: ${2}"
fi
As others have mentioned, chmod does its own validation so it'd probably be best to just use that, but if you absolutely want to do this in a way similar to the code you provided you can do:
#!/bin/sh
if [ $# -eq 0 ]
then
echo "Invalid input"
exit 1
fi
if ! [[ $1 =~ [0-7][0-7][0-7] ]]
then
echo "Invalid input"
exit 1
else
perm=$1
fi
touch ./foo
chmod "${perm}" ./foo
This will ensure that each digit is valid.
What command checks if a directory exists or not within a Bash shell script?
To check if a directory exists:
if [ -d "$DIRECTORY" ]; then
echo "$DIRECTORY does exist."
fi
To check if a directory does not exist:
if [ ! -d "$DIRECTORY" ]; then
echo "$DIRECTORY does not exist."
fi
However, as Jon Ericson points out, subsequent commands may not work as intended if you do not take into account that a symbolic link to a directory will also pass this check.
E.g. running this:
ln -s "$ACTUAL_DIR" "$SYMLINK"
if [ -d "$SYMLINK" ]; then
rmdir "$SYMLINK"
fi
Will produce the error message:
rmdir: failed to remove `symlink': Not a directory
So symbolic links may have to be treated differently, if subsequent commands expect directories:
if [ -d "$LINK_OR_DIR" ]; then
if [ -L "$LINK_OR_DIR" ]; then
# It is a symlink!
# Symbolic link specific commands go here.
rm "$LINK_OR_DIR"
else
# It's a directory!
# Directory command goes here.
rmdir "$LINK_OR_DIR"
fi
fi
Take particular note of the double-quotes used to wrap the variables. The reason for this is explained by 8jean in another answer.
If the variables contain spaces or other unusual characters it will probably cause the script to fail.
Always wrap variables in double quotes when referencing them in a Bash script.
if [ -d "$DIRECTORY" ]; then
# Will enter here if $DIRECTORY exists, even if it contains spaces
fi
Kids these days put spaces and lots of other funny characters in their directory names. (Spaces! Back in my day, we didn't have no fancy spaces!)
One day, one of those kids will run your script with $DIRECTORY set to "My M0viez" and your script will blow up. You don't want that. So use double quotes.
Note the -d test can produce some surprising results:
$ ln -s tmp/ t
$ if [ -d t ]; then rmdir t; fi
rmdir: directory "t": Path component not a directory
File under: "When is a directory not a directory?" The answer: "When it's a symlink to a directory." A slightly more thorough test:
if [ -d t ]; then
if [ -L t ]; then
rm t
else
rmdir t
fi
fi
You can find more information in the Bash manual on Bash conditional expressions and the [ builtin command and the [[ compound commmand.
I find the double-bracket version of test makes writing logic tests more natural:
if [[ -d "${DIRECTORY}" && ! -L "${DIRECTORY}" ]] ; then
echo "It's a bona-fide directory"
fi
Shorter form:
# if $DIR is a directory, then print yes
[ -d "$DIR" ] && echo "Yes"
A simple script to test if a directory or file is present or not:
if [ -d /home/ram/dir ] # For file "if [ -f /home/rama/file ]"
then
echo "dir present"
else
echo "dir not present"
fi
A simple script to check whether the directory is present or not:
mkdir tempdir # If you want to check file use touch instead of mkdir
ret=$?
if [ "$ret" == "0" ]
then
echo "dir present"
else
echo "dir not present"
fi
The above scripts will check if the directory is present or not
$? if the last command is a success it returns "0", else a non-zero value.
Suppose tempdir is already present. Then mkdir tempdir will give an error like below:
mkdir: cannot create directory ‘tempdir’: File exists
To check if a directory exists you can use a simple if structure like this:
if [ -d directory/path to a directory ] ; then
# Things to do
else #if needed #also: elif [new condition]
# Things to do
fi
You can also do it in the negative:
if [ ! -d directory/path to a directory ] ; then
# Things to do when not an existing directory
Note: Be careful. Leave empty spaces on either side of both opening and closing braces.
With the same syntax you can use:
-e: any kind of archive
-f: file
-h: symbolic link
-r: readable file
-w: writable file
-x: executable file
-s: file size greater than zero
You can use test -d (see man test).
-d file True if file exists and is a directory.
For example:
test -d "/etc" && echo Exists || echo Does not exist
Note: The test command is same as conditional expression [ (see: man [), so it's portable across shell scripts.
[ - This is a synonym for the test builtin, but the last argument must, be a literal ], to match the opening [.
For possible options or further help, check:
help [
help test
man test or man [
Or for something completely useless:
[ -d . ] || echo "No"
Here's a very pragmatic idiom:
(cd $dir) || return # Is this a directory,
# and do we have access?
I typically wrap it in a function:
can_use_as_dir() {
(cd ${1:?pathname expected}) || return
}
Or:
assert_dir_access() {
(cd ${1:?pathname expected}) || exit
}
The nice thing about this approach is that I do not have to think of a good error message.
cd will give me a standard one line message to standard error already. It will also give more information than I will be able to provide. By performing the cd inside a subshell ( ... ), the command does not affect the current directory of the caller. If the directory exists, this subshell and the function are just a no-op.
Next is the argument that we pass to cd: ${1:?pathname expected}. This is a more elaborate form of parameter substitution which is explained in more detail below.
Tl;dr: If the string passed into this function is empty, we again exit from the subshell ( ... ) and return from the function with the given error message.
Quoting from the ksh93 man page:
${parameter:?word}
If parameter is set and is non-null then substitute its value;
otherwise, print word and exit from the shell (if not interactive).
If word is omitted then a standard message is printed.
and
If the colon : is omitted from the above expressions, then the
shell only checks whether parameter is set or not.
The phrasing here is peculiar to the shell documentation, as word may refer to any reasonable string, including whitespace.
In this particular case, I know that the standard error message 1: parameter not set is not sufficient, so I zoom in on the type of value that we expect here - the pathname of a directory.
A philosophical note:
The shell is not an object oriented language, so the message says pathname, not directory. At this level, I'd rather keep it simple - the arguments to a function are just strings.
if [ -d "$Directory" -a -w "$Directory" ]
then
#Statements
fi
The above code checks if the directory exists and if it is writable.
More features using find
Check existence of the folder within sub-directories:
found=`find -type d -name "myDirectory"`
if [ -n "$found" ]
then
# The variable 'found' contains the full path where "myDirectory" is.
# It may contain several lines if there are several folders named "myDirectory".
fi
Check existence of one or several folders based on a pattern within the current directory:
found=`find -maxdepth 1 -type d -name "my*"`
if [ -n "$found" ]
then
# The variable 'found' contains the full path where folders "my*" have been found.
fi
Both combinations. In the following example, it checks the existence of the folder in the current directory:
found=`find -maxdepth 1 -type d -name "myDirectory"`
if [ -n "$found" ]
then
# The variable 'found' is not empty => "myDirectory"` exists.
fi
DIRECTORY=/tmp
if [ -d "$DIRECTORY" ]; then
echo "Exists"
fi
Try online
Actually, you should use several tools to get a bulletproof approach:
DIR_PATH=`readlink -f "${the_stuff_you_test}"` # Get rid of symlinks and get abs path
if [[ -d "${DIR_PATH}" ]] ; Then # Now you're testing
echo "It's a dir";
fi
There isn't any need to worry about spaces and special characters as long as you use "${}".
Note that [[]] is not as portable as [], but since most people work with modern versions of Bash (since after all, most people don't even work with command line :-p), the benefit is greater than the trouble.
Have you considered just doing whatever you want to do in the if rather than looking before you leap?
I.e., if you want to check for the existence of a directory before you enter it, try just doing this:
if pushd /path/you/want/to/enter; then
# Commands you want to run in this directory
popd
fi
If the path you give to pushd exists, you'll enter it and it'll exit with 0, which means the then portion of the statement will execute. If it doesn't exist, nothing will happen (other than some output saying the directory doesn't exist, which is probably a helpful side-effect anyways for debugging).
It seems better than this, which requires repeating yourself:
if [ -d /path/you/want/to/enter ]; then
pushd /path/you/want/to/enter
# Commands you want to run in this directory
popd
fi
The same thing works with cd, mv, rm, etc... if you try them on files that don't exist, they'll exit with an error and print a message saying it doesn't exist, and your then block will be skipped. If you try them on files that do exist, the command will execute and exit with a status of 0, allowing your then block to execute.
[[ -d "$DIR" && ! -L "$DIR" ]] && echo "It's a directory and not a symbolic link"
N.B: Quoting variables is a good practice.
Explanation:
-d: check if it's a directory
-L: check if it's a symbolic link
To check more than one directory use this code:
if [ -d "$DIRECTORY1" ] && [ -d "$DIRECTORY2" ] then
# Things to do
fi
Check if the directory exists, else make one:
[ -d "$DIRECTORY" ] || mkdir $DIRECTORY
[ -d ~/Desktop/TEMPORAL/ ] && echo "DIRECTORY EXISTS" || echo "DIRECTORY DOES NOT EXIST"
Using the -e check will check for files and this includes directories.
if [ -e ${FILE_PATH_AND_NAME} ]
then
echo "The file or directory exists."
fi
This answer wrapped up as a shell script
Examples
$ is_dir ~
YES
$ is_dir /tmp
YES
$ is_dir ~/bin
YES
$ mkdir '/tmp/test me'
$ is_dir '/tmp/test me'
YES
$ is_dir /asdf/asdf
NO
# Example of calling it in another script
DIR=~/mydata
if [ $(is_dir $DIR) == "NO" ]
then
echo "Folder doesnt exist: $DIR";
exit;
fi
is_dir
function show_help()
{
IT=$(CAT <<EOF
usage: DIR
output: YES or NO, depending on whether or not the directory exists.
)
echo "$IT"
exit
}
if [ "$1" == "help" ]
then
show_help
fi
if [ -z "$1" ]
then
show_help
fi
DIR=$1
if [ -d $DIR ]; then
echo "YES";
exit;
fi
echo "NO";
As per Jonathan's comment:
If you want to create the directory and it does not exist yet, then the simplest technique is to use mkdir -p which creates the directory — and any missing directories up the path — and does not fail if the directory already exists, so you can do it all at once with:
mkdir -p /some/directory/you/want/to/exist || exit 1
if [ -d "$DIRECTORY" ]; then
# Will enter here if $DIRECTORY exists
fi
This is not completely true...
If you want to go to that directory, you also need to have the execute rights on the directory. Maybe you need to have write rights as well.
Therefore:
if [ -d "$DIRECTORY" ] && [ -x "$DIRECTORY" ] ; then
# ... to go to that directory (even if DIRECTORY is a link)
cd $DIRECTORY
pwd
fi
if [ -d "$DIRECTORY" ] && [ -w "$DIRECTORY" ] ; then
# ... to go to that directory and write something there (even if DIRECTORY is a link)
cd $DIRECTORY
touch foobar
fi
In kind of a ternary form,
[ -d "$directory" ] && echo "exist" || echo "not exist"
And with test:
test -d "$directory" && echo "exist" || echo "not exist"
file="foo"
if [[ -e "$file" ]]; then echo "File Exists"; fi;
The ls command in conjunction with -l (long listing) option returns attributes information about files and directories.
In particular the first character of ls -l output it is usually a d or a - (dash). In case of a d the one listed is a directory for sure.
The following command in just one line will tell you if the given ISDIR variable contains a path to a directory or not:
[[ $(ls -ld "$ISDIR" | cut -c1) == 'd' ]] &&
echo "YES, $ISDIR is a directory." ||
echo "Sorry, $ISDIR is not a directory"
Practical usage:
[claudio#nowhere ~]$ ISDIR="$HOME/Music"
[claudio#nowhere ~]$ ls -ld "$ISDIR"
drwxr-xr-x. 2 claudio claudio 4096 Aug 23 00:02 /home/claudio/Music
[claudio#nowhere ~]$ [[ $(ls -ld "$ISDIR" | cut -c1) == 'd' ]] &&
echo "YES, $ISDIR is a directory." ||
echo "Sorry, $ISDIR is not a directory"
YES, /home/claudio/Music is a directory.
[claudio#nowhere ~]$ touch "empty file.txt"
[claudio#nowhere ~]$ ISDIR="$HOME/empty file.txt"
[claudio#nowhere ~]$ [[ $(ls -ld "$ISDIR" | cut -c1) == 'd' ]] &&
echo "YES, $ISDIR is a directory." ||
echo "Sorry, $ISDIR is not a directoy"
Sorry, /home/claudio/empty file.txt is not a directory
There are great solutions out there, but ultimately every script will fail if you're not in the right directory. So code like this:
if [ -d "$LINK_OR_DIR" ]; then
if [ -L "$LINK_OR_DIR" ]; then
# It is a symlink!
# Symbolic link specific commands go here
rm "$LINK_OR_DIR"
else
# It's a directory!
# Directory command goes here
rmdir "$LINK_OR_DIR"
fi
fi
will execute successfully only if at the moment of execution you're in a directory that has a subdirectory that you happen to check for.
I understand the initial question like this: to verify if a directory exists irrespective of the user's position in the file system. So using the command 'find' might do the trick:
dir=" "
echo "Input directory name to search for:"
read dir
find $HOME -name $dir -type d
This solution is good because it allows the use of wildcards, a useful feature when searching for files/directories. The only problem is that, if the searched directory doesn't exist, the 'find' command will print nothing to standard output (not an elegant solution for my taste) and will have nonetheless a zero exit. Maybe someone could improve on this.
The below find can be used,
find . -type d -name dirname -prune -print
One Liner:
[[ -d $Directory ]] && echo true
(1)
[ -d Piyush_Drv1 ] && echo ""Exists"" || echo "Not Exists"
(2)
[ `find . -type d -name Piyush_Drv1 -print | wc -l` -eq 1 ] && echo Exists || echo "Not Exists"
(3)
[[ -d run_dir && ! -L run_dir ]] && echo Exists || echo "Not Exists"
If an issue is found with one of the approaches provided above:
With the ls command; the cases when a directory does not exists - an error message is shown
[[ `ls -ld SAMPLE_DIR| grep ^d | wc -l` -eq 1 ]] && echo exists || not exists
-ksh: not: not found [No such file or directory]
I am trying to write a script shell that takes two arguments as directory names, and determines if Directory 1 contains Directory 2 or vice versa. And also if there is no relationship between them.
I know the command to check if a directory exists is find -type d, however i was a bit confused as how to check and parse then names. I know i would need if-else loops, just not sure how to check for the conditions?
find won't be needed.
Something similar to this (but not guaranteeing directory name with spaces or some special characters.):
if [ "$dir1" == "$dir2" ]; then
echo "$dir1 == $dir2";
exit;
fi
if grep -E -q "^$dir2" <<< $dir1; then
echo "$dir1 is contained by $dir2."
exit
fi
if grep -E -q "^$dir1" <<< $dir2; then
echo "$dir2 is contained by $dir1.";
fi
However, this does not deal with symbolic links. For example, sym1 -> /usr/local/bin and sym2 -> /usr/local, apparently, sym2 contains sym1.
In addition, this does not deal with strange looking directory names, like /usr/local/./bin, which is the same as /usr/local/bin, or even /usr/local/../bin, which is the same as /usr/bin
--- Update ---
DevSolar metioned that readlink -e can be used to resolve the symbolic link. In my test, it also resolves the strange looking directory names like those with . and ... Thanks to DevSolar.
Do you want this?
if [ -d $1 ];then
a=`find $1 -type d -name $2`
if [ $a ];then
echo "$1 has $2"
else
echo "$1 does NOT has $2"
fi
fi
if [ -d $2 ];then
b=`find $2 -type d -name $1`
if [ $b ];then
echo "$2 has $1"
else
echo "$2 does NOT has $1"
fi
fi
This will do the trick I think,
find -name directory1 |grep directory2
or vice-versa, then use
echo $?
it will give 0 for success and 1 for failure.
In my script, I am asking the user to input a directory and then list all the files in that specific directory. What I want to do with that is to make the display a little better in which I would be able to display a "/" if the item in the directory is another directory and if it is an executable file (not an executable directory), print with a **".
This is what I have:
echo “Directory: “
read thing
for var123 in $thing*
do
echo $var123
done
In a directory I have a few folders and a few scripts that have the execute permission. when I run the script I want to say
/folder1/subfolder1/
/folder1/subfolder2/
/folder1/file1*
/folder1/file2*
I am new to this and have no clue what I am doing. Any help will be greatly appreciated.
You might want to check and make sure the user inputs something that ends in a / first.
e.g.
[[ $thing =~ '/'$ ]] || thing="$thing/"
Also check if it exists
e.g.
[[ -d $thing ]] || exit 1
Then for checking if it's a directory use the -d test as above. To check if executable file use -x. So putting that all together, try:
#!/bin/bash
echo “Directory: “
read thing
[[ $thing =~ '/'$ ]] || thing="$thing/"
[[ -d $thing ]] || exit 1
for var123 in "$thing"*
do
if [[ -f $var123 && -x $var123 ]]; then
echo "$var123**"
elif [[ -d $var123 ]]; then
echo "$var123/"
else
echo "$var123"
fi
done
ls -F is your friend here - if you want to do it for the current directory:
ls -F
If you want to do it for all files & subfolders of the current directory:
find * -exec ls -Fd {} \;
... and for a given directory:
echo "Directory: "
read DIR
find $DIR/* -exec ls -Fd {} \;
Edit: ls -F will append a / to directories and a * to executables. If you want ** instead, just use sed to replace them:
find $DIR/* -exec ls -Fd {} \; | sed 's/\*$/&&/'
And this approach works in all shells, not just bash.
As arguments to my script there are some file paths. Those can, of course, be relative (or contain ~). But for the functions I've written I need paths that are absolute, but do not have their symlinks resolved.
Is there any function for this?
MY_PATH=$(readlink -f $YOUR_ARG) will resolve relative paths like "./" and "../"
Consider this as well (source):
#!/bin/bash
dir_resolve()
{
cd "$1" 2>/dev/null || return $? # cd to desired directory; if fail, quell any error messages but return exit status
echo "`pwd -P`" # output full, link-resolved path
}
# sample usage
if abs_path="`dir_resolve \"$1\"`"
then
echo "$1 resolves to $abs_path"
echo pwd: `pwd` # function forks subshell, so working directory outside function is not affected
else
echo "Could not reach $1"
fi
http://www.linuxquestions.org/questions/programming-9/bash-script-return-full-path-and-filename-680368/page3.html has the following
function abspath {
if [[ -d "$1" ]]
then
pushd "$1" >/dev/null
pwd
popd >/dev/null
elif [[ -e "$1" ]]
then
pushd "$(dirname "$1")" >/dev/null
echo "$(pwd)/$(basename "$1")"
popd >/dev/null
else
echo "$1" does not exist! >&2
return 127
fi
}
which uses pushd/popd to get into a state where pwd is useful.
Simple one-liner:
function abs_path {
(cd "$(dirname '$1')" &>/dev/null && printf "%s/%s" "$PWD" "${1##*/}")
}
Usage:
function do_something {
local file=$(abs_path $1)
printf "Absolute path to %s: %s\n" "$1" "$file"
}
do_something $HOME/path/to/some\ where
I am still trying to figure out how I can get it to be completely oblivious to whether the path exists or not (so it can be used when creating files as well).
This does the trick for me on OS X: $(cd SOME_DIRECTORY 2> /dev/null && pwd -P)
It should work anywhere. The other solutions seemed too complicated.
If your OS supports it, use:
realpath -s "./some/dir"
And using it in a variable:
some_path="$(realpath -s "./some/dir")"
Which will expand your path. Tested on Ubuntu and CentOS, might not be available on yours. Some recommend readlink, but documentation for readlink says:
Note realpath(1) is the preferred command to use for canonicalization functionality.
In case people wonder why I quote my variables, it's to preserve spaces in paths. Like doing realpath some path will give you two different path results. But realpath "some path" will return one. Quoted parameters ftw :)
Thanks to NyanPasu64 for the heads up. You'll want to add -s if you don't want it to follow the symlinks.
Use readlink -f <relative-path>, e.g.
export FULLPATH=`readlink -f ./`
Maybe this is more readable and does not use a subshell and does not change the current dir:
dir_resolve() {
local dir=`dirname "$1"`
local file=`basename "$1"`
pushd "$dir" &>/dev/null || return $? # On error, return error code
echo "`pwd -P`/$file" # output full, link-resolved path with filename
popd &> /dev/null
}
on OS X you can use
stat -f "%N" YOUR_PATH
on linux you might have realpath executable. if not, the following might work (not only for links):
readlink -c YOUR_PATH
There's another method. You can use python embedding in bash script to resolve a relative path.
abs_path=$(python3 - <<END
from pathlib import Path
path = str(Path("$1").expanduser().resolve())
print(path)
END
)
self edit, I just noticed the OP said he's not looking for symlinks resolved:
"But for the functions I've written I need paths that are absolute, but do not have their symlinks resolved."
So guess this isn't so apropos to his question after all. :)
Since I've run into this many times over the years, and this time around I needed a pure bash portable version that I could use on OSX and linux, I went ahead and wrote one:
The living version lives here:
https://github.com/keen99/shell-functions/tree/master/resolve_path
but for the sake of SO, here's the current version (I feel it's well tested..but I'm open to feedback!)
Might not be difficult to make it work for plain bourne shell (sh), but I didn't try...I like $FUNCNAME too much. :)
#!/bin/bash
resolve_path() {
#I'm bash only, please!
# usage: resolve_path <a file or directory>
# follows symlinks and relative paths, returns a full real path
#
local owd="$PWD"
#echo "$FUNCNAME for $1" >&2
local opath="$1"
local npath=""
local obase=$(basename "$opath")
local odir=$(dirname "$opath")
if [[ -L "$opath" ]]
then
#it's a link.
#file or directory, we want to cd into it's dir
cd $odir
#then extract where the link points.
npath=$(readlink "$obase")
#have to -L BEFORE we -f, because -f includes -L :(
if [[ -L $npath ]]
then
#the link points to another symlink, so go follow that.
resolve_path "$npath"
#and finish out early, we're done.
return $?
#done
elif [[ -f $npath ]]
#the link points to a file.
then
#get the dir for the new file
nbase=$(basename $npath)
npath=$(dirname $npath)
cd "$npath"
ndir=$(pwd -P)
retval=0
#done
elif [[ -d $npath ]]
then
#the link points to a directory.
cd "$npath"
ndir=$(pwd -P)
retval=0
#done
else
echo "$FUNCNAME: ERROR: unknown condition inside link!!" >&2
echo "opath [[ $opath ]]" >&2
echo "npath [[ $npath ]]" >&2
return 1
fi
else
if ! [[ -e "$opath" ]]
then
echo "$FUNCNAME: $opath: No such file or directory" >&2
return 1
#and break early
elif [[ -d "$opath" ]]
then
cd "$opath"
ndir=$(pwd -P)
retval=0
#done
elif [[ -f "$opath" ]]
then
cd $odir
ndir=$(pwd -P)
nbase=$(basename "$opath")
retval=0
#done
else
echo "$FUNCNAME: ERROR: unknown condition outside link!!" >&2
echo "opath [[ $opath ]]" >&2
return 1
fi
fi
#now assemble our output
echo -n "$ndir"
if [[ "x${nbase:=}" != "x" ]]
then
echo "/$nbase"
else
echo
fi
#now return to where we were
cd "$owd"
return $retval
}
here's a classic example, thanks to brew:
%% ls -l `which mvn`
lrwxr-xr-x 1 draistrick 502 29 Dec 17 10:50 /usr/local/bin/mvn# -> ../Cellar/maven/3.2.3/bin/mvn
use this function and it will return the -real- path:
%% cat test.sh
#!/bin/bash
. resolve_path.inc
echo
echo "relative symlinked path:"
which mvn
echo
echo "and the real path:"
resolve_path `which mvn`
%% test.sh
relative symlinked path:
/usr/local/bin/mvn
and the real path:
/usr/local/Cellar/maven/3.2.3/libexec/bin/mvn
Do you have to use bash exclusively? I needed to do this and got fed up with differences between Linux and OS X. So I used PHP for a quick and dirty solution.
#!/usr/bin/php <-- or wherever
<?php
{
if($argc!=2)
exit();
$fname=$argv[1];
if(!file_exists($fname))
exit();
echo realpath($fname)."\n";
}
?>
I know it's not a very elegant solution but it does work.