We give the upmost permission to a file or directory, using this command:
sudo chmod -R 777 directory
Now I want to know if this command is already executed for a given directory.
I know I can use -r for read, -w for write, and -x for execution, in [ test ] blocks.
But I want to know two things:
Is it also a directory?
Does it have those permissions for everyone?
How can I get that info?
Update
Based on #Barmar comment, I came up with this. But it's not working:
if [ stat /Temp | grep -oP "(?<=Access: \()[^)]*" == '' ]; then
echo '/Temp folder has full access'
else
sudo chmod -R 777 /Temp
fi
This command works though:
stat /Temp | grep -oP "(?<=Access: \()[^)]*"
# prints => 0777/drwxrwxrwx
How should I fix the syntax error of my if-else statement?
You don't need to process the output of stat with grep; you can ask stat to only produce the specific information you want. See the man page regarding the --format option. We can for example write:
# ask stat for the file type and mode, and put those values into $1
# and $2
set -- $(stat --format '%F %a' /Temp)
if [[ $1 == directory ]]; then
if [[ $2 == 777 ]]; then
echo "/Temp folder has full access"
else
sudo chmod -R 777 /Temp
fi
else
echo "ERROR: /Temp is not a directory!" >&2
fi
A simple example:
#!/bin/bash
function setfullperm(){
[ -d $1 ] && \
(
[ "$(stat --format '%a' $1)" == "777" ] && \
echo "Full permissions are applied." || \
( echo "Setting full permissions" && sudo chmod -R 777 $1 )
) || \
( echo "$1 is not a directory !" && mkdir $1 && setfullperm $1 )
}
export setfullperm
Source the script:
$ source example.sh
Set full permissions (777) on any directory, it tests if the directory exists in the first place, if not it will create it and set the permissions.
It will export the function setfullperm to the shell so you can run it:
>$ setfullperm ali
ali is not a directory !
mkdir: created directory 'ali'
Setting full permissions
>$ setfullperm ali
Full permissions are applied.
If using zsh (But not other shells), you can do it with just a glob pattern:
setopt extended_glob null_glob
if [[ -n /Temp(#q/f777) ]]; then
echo '/Temp folder has full access'
else
sudo chmod -R 777 /Temp
fi
The pattern /Temp(#q/f777) will, with the null_glob and extended_glob options set, expand to an empty string if /Temp is anything but a directory with the exact octal permissions 0777 (And to /Temp if the criteria are met). For more details, see Glob Qualifiers in the zsh manual.
I don't recommend using stat for this. Though widespread, stat isn't POSIX, which means there's no guarantee that your script will work in the future or work on other platforms. If you're writing scripts for a production environment, I'd urge you to consider a different approach.
You're better off using ls(1)'s -l option and passing the file as an argument. From there you can use cut(1)'s -c option to grab the file mode flags.
Get file type:
ls -l <file> | cut -c1
Also, don't forget about test's -d operator, which tests if a file is a directory.
Get owner permissions:
ls -l <file> | cut -c2-4
and so on.
This approach is POSIX compliant and it avoids the shortcomings of using stat.
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 $Directory ]] && echo true
(1)
[ -d Piyush_Drv1 ] && echo ""Exists"" || echo "Not Exists"
(2)
[ `find . -type d -name Piyush_Drv1 -print | wc -l` -eq 1 ] && echo Exists || echo "Not Exists"
(3)
[[ -d run_dir && ! -L run_dir ]] && echo Exists || echo "Not Exists"
If an issue is found with one of the approaches provided above:
With the ls command; the cases when a directory does not exists - an error message is shown
[[ `ls -ld SAMPLE_DIR| grep ^d | wc -l` -eq 1 ]] && echo exists || not exists
-ksh: not: not found [No such file or directory]
I have this simple bash script:
I run ns simulator on each file passed in argument where last argument is some text string to search for.
#!/bin/bash
nsloc="/home/ashish/ns-allinone-2.35/ns-2.35/ns"
temp="temp12345ashish.temp"
j=1
for file in "$#"
do
if [ $j -lt $# ]
then
let j=$j+1
`$nsloc $file > $temp 2>&1`
if grep -l ${BASH_ARGV[0]} $temp
then
echo "$file Successful"
fi
fi
done
I expected:
file1.tcl Successful
I am getting:
temp12345ashish.temp
file1.tcl Successful
When i run the simulator command myself on the terminal i do not get the file name to which output is directed.
I am not getting from where this first line of output is getting printed.
Please explain it.
Thanks in advance.
See man grep, and see specifically the explanation of the -l option.
In your script (above), you are using -l, so grep is telling you (as instructed) the filename where the match occurred.
If you don't want to see the filename, don't use -l, or use -q with it also. Eg:
grep -ql ${BASH_ARGV[0]} $temp
Just silence the grep:
if grep -l ${BASH_ARGV[0]} $temp &> /dev/null