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How do I check if a directory exists in a Bash shell script?

What command checks if a directory exists or not within a Bash shell script?


P
Peter Mortensen

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.


If you want to play it safe with the GNU tools, use of -- is highly recommended (end-of-options marker). Otherwise, if your variable contains something that looks like an option, the script'll fail just as with spaces.
For modern versions of bash, ksh, etc. [...] is a builtin
One thing to keep in mind: [ ! -d "$DIRECTORY" ] will be true either if $DIRECTORY doesn't exist, or if does exist but isn't a directory. Consider something like if [ ! -d "$DIRECTORY" ] ; then mkdir "$DIRECTORY" ; fi; this will fail if "$DIRECTORY" is a file. (Of course you should check whether mkdir succeeded anyway; there are a number of reasons it can fail.)
It might be worth mentioning that as soon as the check has been performed the situation can have changed already due to other processes. In many cases it is better to just create or use the directory and react on a failure.
Instead of testing for both the directory (-d) and the symlink (-L), it's easier just to append a slash to the variable, like if [ -d "${THING:+$THING/}" ]. A directory won't mind the extra slash. A file will evaluate to false. Empty will remain empty, so false. And a symlink will be resolved to its destination. Of course, it depends on your goal. If you want to go there, this is fine. If you want to delete it, then the code in this answer is better.
P
Peter Mortensen

Remember to always wrap variables in double quotes when referencing them in a Bash script. Kids these days grow up with the idea that they can have spaces and lots of other funny characters in their directory names. (Spaces! Back in my days, 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 this.

if [ -d "$DIRECTORY" ]; then
    # Will enter here if $DIRECTORY exists, even if it contains spaces
fi

Another reason to use double quotes is in case $DIRECTORY is not set for some reason.
"always wrap variables in double quotes...in a bash script." For bash, not technically necessary when using [[...]]; see tldp.org/LDP/abs/html/testconstructs.html#DBLBRACKETS (note: no word splitting): "No filename expansion or word splitting takes place between [[ and ]], but there is parameter expansion and command substitution."
Directories on Unix/Linux should not have any whitespaces, and subsequently scripts should not be adapted to it. It's bad enough Windows supports it, with all consequences to Windows scripting, but please, for the love of whatever, no need to introduce unnecessary requirements.
@tvCa I find that users generally prefer to be allowed more flexibility in their directory names rather than being forced to make things easier for developers. (In fact, when dealing with long file names, I find ones without spaces to be a pain as that kills word wrapping even though I myself have suffered in the past from not accounting for paths with spaces in scripts and programs.)
Ha. Spaces are just characters that have no glyphs usually. Anyway, you can escape them with a backslash.
J
Jonathan Leffler

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.


or, assuming it is only necessary to work on directories (and links can be ignored) => if [ -d tmpdir -a ! -L tmpdir ]; then echo "is directory"; rmdir tmpdir; fi ... or, for one command that works on both links & dirs: rm -r tmpdir
J
Jay

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

for if [[ -d "$TARFILE" ]] I'm getting [[: not found
@TheVillageIdiot and @Hedgehog, are you using bash shell? The double bracket isn't universally supported. Here's a SO answer on that point: stackoverflow.com/questions/669452/…
And in Busybox ash with default compilation options [[ ]] is supported, but doesn't in fact provide any different functionality to [ ]. If portability is a concern, stick with [ ] and use the necessary workarounds.
...if using bash constructs in a shell script, the first line of the script should be: #!/bin/bash (and not #!/bin/sh, ksh, etc)
When using double square brackets in bash, you do not need to quote the variables.
L
Lucas

Shorter form:

# if $DIR is a directory, then print yes
[ -d "$DIR" ] && echo "Yes"

Does this work like this: if $dir is a dir, then echo "yes"? A bit of explanation would help :)
cmd && other is a common shorthand for if cmd; then other; fi -- this works with most programming languages which support Boolean logic, and is known as short-circuit evaluation.
The behavior is not the same under set -e (which is a shell programming best practice).
@dolmen the [ -d "$DIR" ] is checked (followed by && echo Yes), so I believe set -e makes no difference to the script behaviour (i.e if the test fails, the script continues normally).
@dolmen That's not correct. The behaviour is exactly the same under set -e. Except that if echo fails, the script will exit. [ -d "$DIR" ] alone (no && cmd ...) will cause the script to exit under set -e, if the test fails (ie. the directory does not exist, or $DIR exists but is not a directory).
J
James Brown

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


The second one would create the directory, if it didn't exist at first. Then it's not idempotent.
👆👆👆. The 2nd one seems dangerous. Since it creates the directory, it's not even true anymore that the dir is not present.
mkdir will also not create the full path (without a param), but the opposite would be even more dangerous, as you are not even able to revert (rm -f) the changes as you don't know which directories it created
P
Peter Mortensen

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

How is this any better than the accepted answer from 2008, aside from going off-topic with the file switches?
better, because of [ ! -d directory/path to a directory ]
k
kenorb

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 [


s
sth

Or for something completely useless:

[ -d . ] || echo "No"

It will never print "No". Current directory always exists, unless deleted by another thread or other ways.
P
Peter Mortensen

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.


This do more than only check for existance: This check for accessibility at your user level. SO question stand for existance only. So right answer is test -d as @Grundlefleck explained.
@F.Hauri - He didn't ask for anything more, that's true. However, I've found that I typically need to know more than that.
And it never occurred to me that no test can be conclusive, unless it runs as root. test -d /unreadable/exists will fail, even if the argument exists.
P
Peter Mortensen
if [ -d "$Directory" -a -w "$Directory" ]
then
    #Statements
fi

The above code checks if the directory exists and if it is writable.


-a is identical in effect to -e. It has been "deprecated," and its use is discouraged.
V
Vishal
DIRECTORY=/tmp

if [ -d "$DIRECTORY" ]; then
    echo "Exists"
fi

Try online


remeber space after [ -> [` `-d. i got error because of missing space
AGAIN, this answer was already given in 2008, with more useful explanations. The only new thing here is the online playground.
f
faho

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


Hi Niel. Your idea may be useful to check the existence of directories depending on a pattern like: find -maxdepth 1 -type d -name 'pattern'. Do you mind if I append in your answer this trick? Cheers ;)
P
Peter Mortensen

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.


P
Peter Mortensen

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.


pushd is to me the most elegant way of doing this. I was about to post it as an answer :)
J
Jahid
[[ -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


An explanation would be in order (by editing your answer, not here in comments).
Quoting variables (even with spaces with paths and metacharacters) is mostly not necessary if you're using Bash's [[...]]. The one exception is the RHS of =, e.g. [[ $VAR = "$VAR" ]]. (This is due to Bash interpreting the RHS as a glob.)
r
rbs

To check more than one directory use this code:

if [ -d "$DIRECTORY1" ] && [ -d "$DIRECTORY2" ] then
    # Things to do
fi

how can you check that it doesn't exists?
P
Peter Mortensen

Check if the directory exists, else make one:

[ -d "$DIRECTORY" ] || mkdir $DIRECTORY

You could use mkdir -p "$DIRECTORY" for the same effect.
Need double quotes around $DIRECTORY in the mkdir part as well. Otherwise, word splitting may result in undesirable results. For example: dir="a b"; mkdir $dir will result in two directories a and b being created, rather than a single a b directory.
d
doubleDown
[ -d ~/Desktop/TEMPORAL/ ] && echo "DIRECTORY EXISTS" || echo "DIRECTORY DOES NOT EXIST"

An explanation would be in order (by editing your answer, not here in comments).
d
doubleDown

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

C
Community

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";

3
3 revs, 3 users 70%

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

P
Peter Mortensen
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

P
Peter Mortensen

In kind of a ternary form,

[ -d "$directory" ] && echo "exist" || echo "not exist"

And with test:

test -d "$directory" && echo "exist" || echo "not exist"

H
Henk Langeveld

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

+1, but it when ISDIR does not exist at all you get an error message as well as your diagnostics message.
a
ajmartin
file="foo" 
if [[ -e "$file" ]]; then echo "File Exists"; fi;

An explanation would be in order (by editing your answer, not here in comments).
P
Peter Mortensen

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.


I'd be offended if a program went looking through my entire hard drive to find a directory rather than just politely looking in my current working directory or using the absolute path I give it. What you've suggested might be nice for a tool named locate but not nice for anything else...
P
Peter Mortensen

The below find can be used,

find . -type d -name dirname -prune -print

P
Peter Mortensen

(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]


What does "above" refer to? The three command lines in this answer or previous answers? (Please respond by editing your answer, not here in comments. Thanks in advance.).
P
Peter Mortensen

Use the file program. Considering all directories are also files in Linux, issuing the following command would suffice:

file $directory_name

Checking a nonexistent file: file blah

Output: cannot open 'blah' (No such file or directory)

Checking an existing directory: file bluh

Output: bluh: directory


Nice idea, however too bad in both cases the file command returns 0 exit code always. Also when the file/directory can not be found :(.