To check if a directory exists in a shell script, you can use the following:
if [ -d "$DIRECTORY" ]; then
# Control will enter here if $DIRECTORY exists.
fi
Or to check if a directory doesn't exist:
if [ ! -d "$DIRECTORY" ]; then
# Control will enter here if $DIRECTORY doesn't exist.
fi
However, as Jon Ericson points out, subsequent commands may not work as intended if you do not take into account that a symbolic link to a directory will also pass this check.
E.g. running this:
ln -s "$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 "$LINK_OR_DIR"
fi
fi
Take particular note of the double-quotes used to wrap the variables. The reason for this is explained by 8jean in another answer.
If the variables contain spaces or other unusual characters it will probably cause the script to fail.
#!/usr/bin/env bash
SCRIPT_DIR="$( cd -- "$( dirname -- "${BASH_SOURCE[0]}" )" &> /dev/null && pwd )"
is a useful one-liner which will give you the full directory name of the script no matter where it is being called from.
It will work as long as the last component of the path used to find the script is not a symlink (directory links are OK). If you also want to resolve any links to the script itself, you need a multi-line solution:
#!/usr/bin/env bash
SOURCE="${BASH_SOURCE[0]}"
while [ -h "$SOURCE" ]; do # resolve $SOURCE until the file is no longer a symlink
DIR="$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )"
SOURCE="$(readlink "$SOURCE")"
[[ $SOURCE != /* ]] && SOURCE="$DIR/$SOURCE" # if $SOURCE was a relative symlink, we need to resolve it relative to the path where the symlink file was located
done
DIR="$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )"
This last one will work with any combination of aliases, source
, bash -c
, symlinks, etc.
Beware: if you cd
to a different directory before running this snippet, the result may be incorrect!
Also, watch out for $CDPATH
gotchas, and stderr output side effects if the user has smartly overridden cd to redirect output to stderr instead (including escape sequences, such as when calling update_terminal_cwd >&2
on Mac). Adding >/dev/null 2>&1
at the end of your cd
command will take care of both possibilities.
To understand how it works, try running this more verbose form:
#!/usr/bin/env bash
SOURCE="${BASH_SOURCE[0]}"
while [ -h "$SOURCE" ]; do # resolve $SOURCE until the file is no longer a symlink
TARGET="$(readlink "$SOURCE")"
if [[ $TARGET == /* ]]; then
echo "SOURCE '$SOURCE' is an absolute symlink to '$TARGET'"
SOURCE="$TARGET"
else
DIR="$( dirname "$SOURCE" )"
echo "SOURCE '$SOURCE' is a relative symlink to '$TARGET' (relative to '$DIR')"
SOURCE="$DIR/$TARGET" # if $SOURCE was a relative symlink, we need to resolve it relative to the path where the symlink file was located
fi
done
echo "SOURCE is '$SOURCE'"
RDIR="$( dirname "$SOURCE" )"
DIR="$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )"
if [ "$DIR" != "$RDIR" ]; then
echo "DIR '$RDIR' resolves to '$DIR'"
fi
echo "DIR is '$DIR'"
And it will print something like:
SOURCE './scriptdir.sh' is a relative symlink to 'sym2/scriptdir.sh' (relative to '.')
SOURCE is './sym2/scriptdir.sh'
DIR './sym2' resolves to '/home/ubuntu/dotfiles/fo fo/real/real1/real2'
DIR is '/home/ubuntu/dotfiles/fo fo/real/real1/real2'
Best Answer
(Usually) The right way
where
${var+x}
is a parameter expansion which evaluates to nothing ifvar
is unset, and substitutes the stringx
otherwise.Quotes Digression
Quotes can be omitted (so we can say
${var+x}
instead of"${var+x}"
) because this syntax & usage guarantees this will only expand to something that does not require quotes (since it either expands tox
(which contains no word breaks so it needs no quotes), or to nothing (which results in[ -z ]
, which conveniently evaluates to the same value (true) that[ -z "" ]
does as well)).However, while quotes can be safely omitted, and it was not immediately obvious to all (it wasn't even apparent to the first author of this quotes explanation who is also a major Bash coder), it would sometimes be better to write the solution with quotes as
[ -z "${var+x}" ]
, at the very small possible cost of an O(1) speed penalty. The first author also added this as a comment next to the code using this solution giving the URL to this answer, which now also includes the explanation for why the quotes can be safely omitted.(Often) The wrong way
This is often wrong because it doesn't distinguish between a variable that is unset and a variable that is set to the empty string. That is to say, if
var=''
, then the above solution will output "var is blank".The distinction between unset and "set to the empty string" is essential in situations where the user has to specify an extension, or additional list of properties, and that not specifying them defaults to a non-empty value, whereas specifying the empty string should make the script use an empty extension or list of additional properties.
The distinction may not be essential in every scenario though. In those cases
[ -z "$var" ]
will be just fine.