This checks if a file exists:
#!/bin/bash
FILE=$1
if [ -f $FILE ]; then
echo "File $FILE exists."
else
echo "File $FILE does not exist."
fi
How do I only check if the file does not exist?
The test command (written as [ here) has a "not" logical operator, ! (exclamation mark):
if [ ! -f /tmp/foo.txt ]; then
echo "File not found!"
fi
Bash File Testing
-b filename - Block special file
-c filename - Special character file
-d directoryname - Check for directory Existence
-e filename - Check for file existence, regardless of type (node, directory, socket, etc.)
-f filename - Check for regular file existence not a directory
-G filename - Check if file exists and is owned by effective group ID
-G filename set-group-id - True if file exists and is set-group-id
-k filename - Sticky bit
-L filename - Symbolic link
-O filename - True if file exists and is owned by the effective user id
-r filename - Check if file is a readable
-S filename - Check if file is socket
-s filename - Check if file is nonzero size
-u filename - Check if file set-user-id bit is set
-w filename - Check if file is writable
-x filename - Check if file is executable
How to use:
#!/bin/bash
file=./file
if [ -e "$file" ]; then
echo "File exists"
else
echo "File does not exist"
fi
A test expression can be negated by using the ! operator
#!/bin/bash
file=./file
if [ ! -e "$file" ]; then
echo "File does not exist"
else
echo "File exists"
fi
Negate the expression inside test (for which [ is an alias) using !:
#!/bin/bash
FILE=$1
if [ ! -f "$FILE" ]
then
echo "File $FILE does not exist"
fi
The relevant man page is man test or, equivalently, man [ -- or help test or help [ for the built-in bash command.
Alternatively (less commonly used) you can negate the result of test using:
if ! [ -f "$FILE" ]
then
echo "File $FILE does not exist"
fi
That syntax is described in "man 1 bash" in sections "Pipelines" and "Compound Commands".
[[ -f $FILE ]] || printf '%s does not exist!\n' "$FILE"
Also, it's possible that the file is a broken symbolic link, or a non-regular file, like e.g. a socket, device or fifo. For example, to add a check for broken symlinks:
if [[ ! -f $FILE ]]; then
if [[ -L $FILE ]]; then
printf '%s is a broken symlink!\n' "$FILE"
else
printf '%s does not exist!\n' "$FILE"
fi
fi
It's worth mentioning that if you need to execute a single command you can abbreviate
if [ ! -f "$file" ]; then
echo "$file"
fi
to
test -f "$file" || echo "$file"
or
[ -f "$file" ] || echo "$file"
I prefer to do the following one-liner, in POSIX shell compatible format:
$ [ -f "/$DIR/$FILE" ] || echo "$FILE NOT FOUND"
$ [ -f "/$DIR/$FILE" ] && echo "$FILE FOUND"
For a couple of commands, like I would do in a script:
$ [ -f "/$DIR/$FILE" ] || { echo "$FILE NOT FOUND" ; exit 1 ;}
Once I started doing this, I rarely use the fully typed syntax anymore!!
To test file existence, the parameter can be any one of the following:
-e: Returns true if file exists (regular file, directory, or symlink)
-f: Returns true if file exists and is a regular file
-d: Returns true if file exists and is a directory
-h: Returns true if file exists and is a symlink
All the tests below apply to regular files, directories, and symlinks:
-r: Returns true if file exists and is readable
-w: Returns true if file exists and is writable
-x: Returns true if file exists and is executable
-s: Returns true if file exists and has a size > 0
Example script:
#!/bin/bash
FILE=$1
if [ -f "$FILE" ]; then
echo "File $FILE exists"
else
echo "File $FILE does not exist"
fi
You can do this:
[[ ! -f "$FILE" ]] && echo "File doesn't exist"
or
if [[ ! -f "$FILE" ]]; then
echo "File doesn't exist"
fi
If you want to check for file and folder both, then use -e option instead of -f. -e returns true for regular files, directories, socket, character special files, block special files etc.
You should be careful about running test for an unquoted variable, because it might produce unexpected results:
$ [ -f ]
$ echo $?
0
$ [ -f "" ]
$ echo $?
1
The recommendation is usually to have the tested variable surrounded by double quotation marks:
#!/bin/sh
FILE=$1
if [ ! -f "$FILE" ]
then
echo "File $FILE does not exist."
fi
In
[ -f "$file" ]
the [ command does a stat() (not lstat()) system call on the path stored in $file and returns true if that system call succeeds and the type of the file as returned by stat() is "regular".
So if [ -f "$file" ] returns true, you can tell the file does exist and is a regular file or a symlink eventually resolving to a regular file (or at least it was at the time of the stat()).
However if it returns false (or if [ ! -f "$file" ] or ! [ -f "$file" ] return true), there are many different possibilities:
the file doesn't exist
the file exists but is not a regular file (could be a device, fifo, directory, socket...)
the file exists but you don't have search permission to the parent directory
the file exists but that path to access it is too long
the file is a symlink to a regular file, but you don't have search permission to some of the directories involved in the resolution of the symlink.
... any other reason why the stat() system call may fail.
In short, it should be:
if [ -f "$file" ]; then
printf '"%s" is a path to a regular file or symlink to regular file\n' "$file"
elif [ -e "$file" ]; then
printf '"%s" exists but is not a regular file\n' "$file"
elif [ -L "$file" ]; then
printf '"%s" exists, is a symlink but I cannot tell if it eventually resolves to an actual file, regular or not\n' "$file"
else
printf 'I cannot tell if "%s" exists, let alone whether it is a regular file or not\n' "$file"
fi
To know for sure that the file doesn't exist, we'd need the stat() system call to return with an error code of ENOENT (ENOTDIR tells us one of the path components is not a directory is another case where we can tell the file doesn't exist by that path). Unfortunately the [ command doesn't let us know that. It will return false whether the error code is ENOENT, EACCESS (permission denied), ENAMETOOLONG or anything else.
The [ -e "$file" ] test can also be done with ls -Ld -- "$file" > /dev/null. In that case, ls will tell you why the stat() failed, though the information can't easily be used programmatically:
$ file=/var/spool/cron/crontabs/root
$ if [ ! -e "$file" ]; then echo does not exist; fi
does not exist
$ if ! ls -Ld -- "$file" > /dev/null; then echo stat failed; fi
ls: cannot access '/var/spool/cron/crontabs/root': Permission denied
stat failed
At least ls tells me it's not because the file doesn't exist that it fails. It's because it can't tell whether the file exists or not. The [ command just ignored the problem.
With the zsh shell, you can query the error code with the $ERRNO special variable after the failing [ command, and decode that number using the $errnos special array in the zsh/system module:
zmodload zsh/system
ERRNO=0
if [ ! -f "$file" ]; then
err=$ERRNO
case $errnos[err] in
("") echo exists, not a regular file;;
(ENOENT|ENOTDIR)
if [ -L "$file" ]; then
echo broken link
else
echo does not exist
fi;;
(*) syserror -p "can't tell: " "$err"
esac
fi
(beware the $errnos support was broken with some versions of zsh when built with recent versions of gcc).
There are three distinct ways to do this:
Negate the exit status with bash (no other answer has said this):
if ! [ -e "$file" ]; then
echo "file does not exist"
fi
Or:
! [ -e "$file" ] && echo "file does not exist"
Negate the test inside the test command [ (that is the way most answers before have presented):
if [ ! -e "$file" ]; then
echo "file does not exist"
fi
Or:
[ ! -e "$file" ] && echo "file does not exist"
Act on the result of the test being negative (|| instead of &&):
Only:
[ -e "$file" ] || echo "file does not exist"
This looks silly (IMO), don't use it unless your code has to be portable to the Bourne shell (like the /bin/sh of Solaris 10 or earlier) that lacked the pipeline negation operator (!):
if [ -e "$file" ]; then
:
else
echo "file does not exist"
fi
envfile=.env
if [ ! -f "$envfile" ]
then
echo "$envfile does not exist"
exit 1
fi
To reverse a test, use "!".
That is equivalent to the "not" logical operator in other languages. Try this:
if [ ! -f /tmp/foo.txt ];
then
echo "File not found!"
fi
Or written in a slightly different way:
if [ ! -f /tmp/foo.txt ]
then echo "File not found!"
fi
Or you could use:
if ! [ -f /tmp/foo.txt ]
then echo "File not found!"
fi
Or, presing all together:
if ! [ -f /tmp/foo.txt ]; then echo "File not found!"; fi
Which may be written (using then "and" operator: &&) as:
[ ! -f /tmp/foo.txt ] && echo "File not found!"
Which looks shorter like this:
[ -f /tmp/foo.txt ] || echo "File not found!"
The test thing may count too. It worked for me (based on Bash Shell: Check File Exists or Not):
test -e FILENAME && echo "File exists" || echo "File doesn't exist"
This code also working .
#!/bin/bash
FILE=$1
if [ -f $FILE ]; then
echo "File '$FILE' Exists"
else
echo "The File '$FILE' Does Not Exist"
fi
The simplest way
FILE=$1
[ ! -e "${FILE}" ] && echo "does not exist" || echo "exists"
This shell script also works for finding a file in a directory:
echo "enter file"
read -r a
if [ -s /home/trainee02/"$a" ]
then
echo "yes. file is there."
else
echo "sorry. file is not there."
fi
sometimes it may be handy to use && and || operators.
Like in (if you have command "test"):
test -b $FILE && echo File not there!
or
test -b $FILE || echo File there!
If you want to use test instead of [], then you can use ! to get the negation:
if ! test "$FILE"; then
echo "does not exist"
fi
You can also group multiple commands in the one liner
[ -f "filename" ] || ( echo test1 && echo test2 && echo test3 )
or
[ -f "filename" ] || { echo test1 && echo test2 && echo test3 ;}
If filename doesn't exit, the output will be
test1
test2
test3
Note: ( ... ) runs in a subshell, { ... ;} runs in the same shell.
I´m trying to make a script which takes a single command-line argument. Then it looks on the argument and if its a directory name, it just prints that this directory exists. If its a file name, it prints out the file exists. Otherwise, it tries to create a directory with this name and tests whether it was successful and reports this on the standard output.
my code is:
while read argument; do
if [ $argument -d ]; then
echo "Directory exists"
elif [ $argument -e ]
echo "File exists"
else
mkdir $argument
if [ $argument -d]; then
echo "Directory was created"
else
echo "Error while creating the directory"
fi
fi
done
Then I run the code ./file_name.sh argument. If I run the code like this, I get an error on line 8, which is just "else". While is probably not necessary here, it was the first option how to accept an argument from the command line that came to my mind.
As you mentioned, you need single command line argument, So no need to loop
#!/bin/bash
if [[ -z "$1" ]]; then
echo "Help : You have to pass one argument"
exit 0
fi
if [[ -d "$1" ]]; then
echo "Directory exists"
elif [[ -f "$1" ]]; then
echo "File exists"
else
mkdir "$1"
if [ -d "$1" ]; then
echo "Directory was created"
else
echo "Error while creating the directory"
fi
fi
if [ -d $1 ]; then
echo "Directory exists"
elif [ -e $1 ]; then
echo "File exists"
else
mkdir $1
if [ -d $1 ]; then
echo "Directory was created"
else
echo "Error while creating the directory"
fi
fi
I made up this solution thanks to the links provided, thank you.
This is a part of a script I'm trying to write.
If the directory exist there is no problem, but I don't know what to do if it not exists.
Some one has an idea how to solve this.
Thanks
do echo "$number"
newdir="../FILE-ID/*/${number:2:1}${number:1:1}/+33$number"
nbrdir=$(ls -lrtd $newdir|wc -l)
echo "$nbrdir"
if [ "$nbrdir" -gt 1 ]; then
echo "$number"
echo "error 1.greater"
fi
if [ "$nbrdir" -eq 1 ]; then
echo " equal 1"
else
echo "equal 0"
fi
done
You can use the following to test whether a file (including a directory) exists:
if [[ -f ${newdir} ]]; then ...
or another switch if you want to test specifically just for a directory:
if [[ -d ${newdir} ]]; then ...
A more comprehensive list can be found here.
If your directory name contains wildcards and matches potentially zero or more, you can use:
nbrdir=$(ls -1d ${newdir} 2>/dev/null | wc -l)
This should give you a count of matching directories. Note that any error message which might be generated will be directed to /dev/null which is probably what you want here to avoid errors messing up your output.
[ -d $newdir ] && echo "Directory exists" || echo "Directory does not exists"
If a file called “output” already exists, rather than a directory, the script
should display an error and quit.
here is my code so far
for file in *
do
if [ ! -f output ]
then echo "error"
exit 1
fi
done
for file in *; do
if [ "$file" = "output" -a -f "$file" ]; then
echo "error"
exit 1
fi
done
Or
for file in *; do
if [ "$file" = "output" ] && [ -f "$file" ]; then
echo "error"
exit 1
fi
done
And with bash, this one's preferred:
for file in *; do
if [[ $file == output && -f $file ]]; then
echo "error"
exit 1
fi
done
If you want to check if the filename contains the word, not just exactly matches it:
for file in *; do
if [[ $file == *output* && -f $file ]]; then
echo "error"
exit 1
fi
done
Why are we processing every file in the subdirectory? Very Odd.
if [ -f output ]; then
echo "'output exists and is a file"
exit 1
fi
The test command (which is also [) (and is also built-in to most shells (see bash man page too) ), responds with a TRUE response for -f output only when output is a file. You can check if it's a directory with -d.
touch something
if [ -f something ]; then echo "something is a file"; fi
if [ -d something ]; then echo "something is not a file"; fi
rm something
mkdir something
if [ -f something ]; then echo "something is not a subdir"; fi
if [ -d something ]; then echo "something is a subdir"; fi
rmdir something
If you try those commands, you'll get:
something is a file
something is a subdir
No point in iterating through the entire directory contents if you're just looking if a specific file/dir exists.
This works just fine
if [[ -e img.png ]]
then
echo "exist"
else
echo "doesn't exist"
fi
but what if I know that there might be imgage with name img but I do not know if the file is .jpg , .gif , .jpeg , .tff and so on.
I do not care what is the extension I just want to know if there is a file with name 'img'
How can I do this ?
You can do:
files=$(ls img.* 2> /dev/null | wc -l)
if [ "$files" != "0" ]
then
echo "exist"
else
echo "doesn't exist"
fi
You can use the following scripts
files=`ls img.* 2>/dev/null`
if [ "$files" -a ${#files[#]} ]; then
echo "exist"
else
echo "doesn't exist"
fi
In this snippet, you use ls img.* to list all the files in current working directory whose name match the pattern img.*.
The result is stored into an array named files.
Then check size of the array to determine whether required files exist.
See this for how to get length of the array.
Something like this should do the job:
if [[ $(ls img.*) ]]; then
echo "file exist";
else
echo "file does not exist";
fi
I recommend to have a look at bash's pattern matching:
http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_04_03.html
Without any external command:
$ for i in img.*
> do
> [ -f $i ] && echo exist || echo not exist
> break
> done
Check if any file is present. If present print exist , else not exist, and break immediately. The "-f" check is needed because if no files are present, still the loop runs once with i as "img.*" itself.
shopt -s nullglob
files=( img.* )
if (( ${#files[#]} == 0 )); then
echo "there are no 'img' files"
fi
If you don't use nullglob then, if there are no such files, the array will have 1 element, the literal string "img.*".