Issues in getting $PWD in shell script - bash

I am calling a system call from my linux application.
/* Some file.c */
file.c is embedded in an executable called file.elf. this file.elf is present in directory
/home/ubuntu/file.elf
when i execute the file.elf the echo $BB_PATH prints the executable directory path. i am expecting the directory path to be the path where the script has been placed. i.e
/home/ubuntu/Desktop/BIN/BB/Chk_File.sh
How can this be acheived ?
if(!(system("ls /home/ubuntu/Desktop/BIN/BB")))
{
/* Test Path : remove after testing */
dw_flag = system("/home/ubuntu/Desktop/BIN/BB/Chk_File.sh");//Call to execute Script
dw_flag = WEXITSTATUS(dw_flag);
}
this in turns call the file CHK_File.sh
ret_val=0
BB_PATH=$(pwd)
echo $BB_PATH
if [ ! -f ACTION_TAG.txt ]
then
echo " ACTION_TAG NOT PRESENT "
else
ret_val=1
fi
echo $ret_val
exit $ret_val

You can use dirname to get the directory of the shell script.
echo `dirname $0`

Adding this below lines in script helped me
if [ -L $0 ] ; then
DIR=$(dirname $(readlink -f $0)) ;
else
DIR=$(dirname $0) ;
fi ;
echo $DIR
Now we dont need to worry from which directory the script is called !! Great !!

Related

How do I reference the current directory inside a terminal command file? [duplicate]

How can I determine the name of the Bash script file inside the script itself?
Like if my script is in file runme.sh, then how would I make it to display "You are running runme.sh" message without hardcoding that?
me=`basename "$0"`
For reading through a symlink1, which is usually not what you want (you usually don't want to confuse the user this way), try:
me="$(basename "$(test -L "$0" && readlink "$0" || echo "$0")")"
IMO, that'll produce confusing output. "I ran foo.sh, but it's saying I'm running bar.sh!? Must be a bug!" Besides, one of the purposes of having differently-named symlinks is to provide different functionality based on the name it's called as (think gzip and gunzip on some platforms).
1 That is, to resolve symlinks such that when the user executes foo.sh which is actually a symlink to bar.sh, you wish to use the resolved name bar.sh rather than foo.sh.
# ------------- SCRIPT ------------- #
#!/bin/bash
echo
echo "# arguments called with ----> ${#} "
echo "# \$1 ----------------------> $1 "
echo "# \$2 ----------------------> $2 "
echo "# path to me ---------------> ${0} "
echo "# parent path --------------> ${0%/*} "
echo "# my name ------------------> ${0##*/} "
echo
exit
# ------------- CALLED ------------- #
# Notice on the next line, the first argument is called within double,
# and single quotes, since it contains two words
$ /misc/shell_scripts/check_root/show_parms.sh "'hello there'" "'william'"
# ------------- RESULTS ------------- #
# arguments called with ---> 'hello there' 'william'
# $1 ----------------------> 'hello there'
# $2 ----------------------> 'william'
# path to me --------------> /misc/shell_scripts/check_root/show_parms.sh
# parent path -------------> /misc/shell_scripts/check_root
# my name -----------------> show_parms.sh
# ------------- END ------------- #
With bash >= 3 the following works:
$ ./s
0 is: ./s
BASH_SOURCE is: ./s
$ . ./s
0 is: bash
BASH_SOURCE is: ./s
$ cat s
#!/bin/bash
printf '$0 is: %s\n$BASH_SOURCE is: %s\n' "$0" "$BASH_SOURCE"
$BASH_SOURCE gives the correct answer when sourcing the script.
This however includes the path so to get the scripts filename only, use:
$(basename $BASH_SOURCE)
If the script name has spaces in it, a more robust way is to use "$0" or "$(basename "$0")" - or on MacOS: "$(basename \"$0\")". This prevents the name from getting mangled or interpreted in any way. In general, it is good practice to always double-quote variable names in the shell.
If you want it without the path then you would use ${0##*/}
To answer Chris Conway, on Linux (at least) you would do this:
echo $(basename $(readlink -nf $0))
readlink prints out the value of a symbolic link. If it isn't a symbolic link, it prints the file name. -n tells it to not print a newline. -f tells it to follow the link completely (if a symbolic link was a link to another link, it would resolve that one as well).
I've found this line to always work, regardless of whether the file is being sourced or run as a script.
echo "${BASH_SOURCE[${#BASH_SOURCE[#]} - 1]}"
If you want to follow symlinks use readlink on the path you get above, recursively or non-recursively.
The reason the one-liner works is explained by the use of the BASH_SOURCE environment variable and its associate FUNCNAME.
BASH_SOURCE
An array variable whose members are the source filenames where the corresponding shell function names in the FUNCNAME array variable are defined. The shell function ${FUNCNAME[$i]} is defined in the file ${BASH_SOURCE[$i]} and called from ${BASH_SOURCE[$i+1]}.
FUNCNAME
An array variable containing the names of all shell functions currently in the execution call stack. The element with index 0 is the name of any currently-executing shell function. The bottom-most element (the one with the highest index) is "main". This variable exists only when a shell function is executing. Assignments to FUNCNAME have no effect and return an error status. If FUNCNAME is unset, it loses its special properties, even if it is subsequently reset.
This variable can be used with BASH_LINENO and BASH_SOURCE. Each element of FUNCNAME has corresponding elements in BASH_LINENO and BASH_SOURCE to describe the call stack. For instance, ${FUNCNAME[$i]} was called from the file ${BASH_SOURCE[$i+1]} at line number ${BASH_LINENO[$i]}. The caller builtin displays the current call stack using this information.
[Source: Bash manual]
Since some comments asked about the filename without extension, here's an example how to accomplish that:
FileName=${0##*/}
FileNameWithoutExtension=${FileName%.*}
Enjoy!
These answers are correct for the cases they state but there is a still a problem if you run the script from another script using the 'source' keyword (so that it runs in the same shell). In this case, you get the $0 of the calling script. And in this case, I don't think it is possible to get the name of the script itself.
This is an edge case and should not be taken TOO seriously. If you run the script from another script directly (without 'source'), using $0 will work.
Re: Tanktalus's (accepted) answer above, a slightly cleaner way is to use:
me=$(readlink --canonicalize --no-newline $0)
If your script has been sourced from another bash script, you can use:
me=$(readlink --canonicalize --no-newline $BASH_SOURCE)
I agree that it would be confusing to dereference symlinks if your objective is to provide feedback to the user, but there are occasions when you do need to get the canonical name to a script or other file, and this is the best way, imo.
this="$(dirname "$(realpath "$BASH_SOURCE")")"
This resolves symbolic links (realpath does that), handles spaces (double quotes do this), and will find the current script name even when sourced (. ./myscript) or called by other scripts ($BASH_SOURCE handles that). After all that, it is good to save this in a environment variable for re-use or for easy copy elsewhere (this=)...
You can use $0 to determine your script name (with full path) - to get the script name only you can trim that variable with
basename $0
if your invoke shell script like
/home/mike/runme.sh
$0 is full name
/home/mike/runme.sh
basename $0 will get the base file name
runme.sh
and you need to put this basic name into a variable like
filename=$(basename $0)
and add your additional text
echo "You are running $filename"
so your scripts like
/home/mike/runme.sh
#!/bin/bash
filename=$(basename $0)
echo "You are running $filename"
This works fine with ./self.sh, ~/self.sh, source self.sh, source ~/self.sh:
#!/usr/bin/env bash
self=$(readlink -f "${BASH_SOURCE[0]}")
basename=$(basename "$self")
echo "$self"
echo "$basename"
Credits: I combined multiple answers to get this one.
echo "$(basename "`test -L ${BASH_SOURCE[0]} \
&& readlink ${BASH_SOURCE[0]} \
|| echo ${BASH_SOURCE[0]}`")"
In bash you can get the script file name using $0. Generally $1, $2 etc are to access CLI arguments. Similarly $0 is to access the name which triggers the script(script file name).
#!/bin/bash
echo "You are running $0"
...
...
If you invoke the script with path like /path/to/script.sh then $0 also will give the filename with path. In that case need to use $(basename $0) to get only script file name.
Short, clear and simple, in my_script.sh
#!/bin/bash
running_file_name=$(basename "$0")
echo "You are running '$running_file_name' file."
Out put:
./my_script.sh
You are running 'my_script.sh' file.
Info thanks to Bill Hernandez. I added some preferences I'm adopting.
#!/bin/bash
function Usage(){
echo " Usage: show_parameters [ arg1 ][ arg2 ]"
}
[[ ${#2} -eq 0 ]] && Usage || {
echo
echo "# arguments called with ----> ${#} "
echo "# \$1 -----------------------> $1 "
echo "# \$2 -----------------------> $2 "
echo "# path to me ---------------> ${0} " | sed "s/$USER/\$USER/g"
echo "# parent path --------------> ${0%/*} " | sed "s/$USER/\$USER/g"
echo "# my name ------------------> ${0##*/} "
echo
}
Cheers
DIRECTORY=$(cd `dirname $0` && pwd)
I got the above from another Stack Overflow question, Can a Bash script tell what directory it's stored in?, but I think it's useful for this topic as well.
Here is what I came up with, inspired by Dimitre Radoulov's answer (which I upvoted, by the way).
script="$BASH_SOURCE"
[ -z "$BASH_SOURCE" ] && script="$0"
echo "Called $script with $# argument(s)"
regardless of the way you call your script
. path/to/script.sh
or
./path/to/script.sh
$0 will give the name of the script you are running. Create a script file and add following code
#!/bin/bash
echo "Name of the file is $0"
then run from terminal like this
./file_name.sh
To get the "realpath" of script or sourced scripts in all cases :
fullname=$(readlink $0) # Take care of symbolic links
dirname=${fullname%/*} # Get (most of the time) the dirname
realpath=$(dirname $BASH_SOURCE) # TO handle sourced scripts
[ "$realpath" = '.' ] && realpath=${dirname:-.}
Here is the bash script to generate (in a newly created "workdir" subdir and in "mytest" in current dir), a bash script which in turn will source another script, which in turm will call a bash defined function .... tested with many ways to launch them :
#!/bin/bash
##############################################################
ret=0
fullname=$(readlink $0) # Take care of symbolic links
dirname=${fullname%/*} # Get (most of the time) the dirname
realpath=$(dirname $BASH_SOURCE) # TO handle sourced scripts
[ "$realpath" = '.' ] && realpath=${dirname:-.}
fullname_withoutextension=${fullname%.*}
mkdir -p workdir
cat <<'EOD' > workdir/_script_.sh
#!/bin/bash
##############################################################
ret=0
fullname=$(readlink $0) # Take care of symbolic links
dirname=${fullname%/*} # Get (most of the time) the dirname
realpath=$(dirname $BASH_SOURCE) # TO handle sourced scripts
[ "$realpath" = '.' ] && realpath=${dirname:-.}
fullname_withoutextension=${fullname%.*}
echo
echo "# ------------- RESULTS ------------- #"
echo "# path to me (\$0)-----------> ${0} "
echo "# arguments called with ----> ${#} "
echo "# \$1 -----------------------> $1 "
echo "# \$2 -----------------------> $2 "
echo "# path to me (\$fullname)----> ${fullname} "
echo "# parent path(\${0%/*})------> ${0%/*} "
echo "# parent path(\$dirname)-----> ${dirname} "
echo "# my name ----\${0##*/}------> ${0##*/} "
echo "# my source -\${BASH_SOURCE}-> ${BASH_SOURCE} "
echo "# parent path(from BASH_SOURCE) -> $(dirname $BASH_SOURCE)"
echo "# my function name -\${FUNCNAME[0]}------> ${FUNCNAME[0]}"
echo "# my source or script real path (realpath)------------------> $realpath"
echo
[ "$realpath" = "workdir" ] || ret=1
[ $ret = 0 ] || echo "*******************************************************"
[ $ret = 0 ] || echo "*********** ERROR **********************************"
[ $ret = 0 ] || echo "*******************************************************"
show_params () {
echo
echo "# --- RESULTS FROM show_params() ---- #"
echo "# path to me (\$0)-----------> ${0} "
echo "# arguments called with ----> ${#} "
echo "# \$1 -----------------------> $1 "
echo "# \$2 -----------------------> $2 "
echo "# path to me (\$fullname)----> ${fullname} "
echo "# parent path(\${0%/*})------> ${0%/*} "
echo "# parent path(\$dirname)-----> ${dirname} "
echo "# my name ----\${0##*/}------> ${0##*/} "
echo "# my source -\${BASH_SOURCE}-> ${BASH_SOURCE} "
echo "# parent path(from BASH_SOURCE) -> $(dirname $BASH_SOURCE)"
echo "# my function name -\${FUNCNAME[0]}------> ${FUNCNAME[0]}"
echo "# my source or script real path (realpath)------------------> $realpath"
echo
[ "$realpath" = "workdir" ] || ret=1
[ $ret = 0 ] || echo "*******************************************************"
[ $ret = 0 ] || echo "*********** ERROR **********************************"
[ $ret = 0 ] || echo "*******************************************************"
}
show_params "$#"
EOD
cat workdir/_script_.sh > workdir/_side_by_side_script_sourced.inc
cat <<'EOD' >> workdir/_script_.sh
echo "# . $realpath/_side_by_side_script_sourced.inc 'hello there' 'william'"
. $realpath/_side_by_side_script_sourced.inc 'hello there' 'william'
[ $ret = 0 ] || echo "*******************************************************"
[ $ret = 0 ] || echo "*********** ERROR **********************************"
[ $ret = 0 ] || echo "*******************************************************"
EOD
chmod +x workdir/_script_.sh
[ -L _mytest_ ] && rm _mytest_
ln -s workdir/_script_.sh _mytest_
# ------------- CALLED ------------- #
called_by () {
echo '=========================================================================='
echo " Called by : " "$#"
echo '=========================================================================='
eval "$#"
}
called_by bash _mytest_
called_by ./_mytest_
called_by bash workdir/_script_.sh
called_by workdir/_script_.sh
called_by . workdir/_script_.sh
# ------------- RESULTS ------------- #
echo
echo
[ $ret = 0 ] || echo "*******************************************************"
[ $ret = 0 ] || echo "*********** ERROR **********************************"
[ $ret = 0 ] || echo "*******************************************************"
echo
[ $ret = 0 ] && echo ".... location of scripts (\$realpath) should always be equal to $realpath, for all test cases at date".
echo
# ------------- END ------------- #
echo "You are running $0"
somthing like this?
export LC_ALL=en_US.UTF-8
#!/bin/bash
#!/bin/sh
#----------------------------------------------------------------------
start_trash(){
ver="htrash.sh v0.0.4"
$TRASH_DIR # url to trash $MY_USER
$TRASH_SIZE # Show Trash Folder Size
echo "Would you like to empty Trash [y/n]?"
read ans
if [ $ans = y -o $ans = Y -o $ans = yes -o $ans = Yes -o $ans = YES ]
then
echo "'yes'"
cd $TRASH_DIR && $EMPTY_TRASH
fi
if [ $ans = n -o $ans = N -o $ans = no -o $ans = No -o $ans = NO ]
then
echo "'no'"
fi
return $TRUE
}
#-----------------------------------------------------------------------
start_help(){
echo "HELP COMMANDS-----------------------------"
echo "htest www open a homepage "
echo "htest trash empty trash "
return $TRUE
} #end Help
#-----------------------------------------------#
homepage=""
return $TRUE
} #end cpdebtemp
# -Case start
# if no command line arg given
# set val to Unknown
if [ -z $1 ]
then
val="*** Unknown ***"
elif [ -n $1 ]
then
# otherwise make first arg as val
val=$1
fi
# use case statement to make decision for rental
case $val in
"trash") start_trash ;;
"help") start_help ;;
"www") firefox $homepage ;;
*) echo "Sorry, I can not get a $val for you!";;
esac
# Case stop

How to rename the filename of that folder name and copy all files into one folder

I am new to unix . I got a requirement like this .
I have xml folder in the server . In that folder , everyday , i will get different employee details for each employee in one folder .
/server/user/home/xml/e1100123/Employeedetails.xml
/server/user/home/xml/e1100123/Employeesalary.xml
/server/user/home/xml/e1100123/Employeeleaves.xml
/server/user/home/xml/e1100123/Employeestatus.xml
/server/user/home/xml/e1100155/Employeedetails.xml
/server/user/home/xml/e1100155/Employeesalary.xml
/server/user/home/xml/e1100155/Employeeleaves.xml
/server/user/home/xml/e1100155/Employeestatus.xml
I have to group all employees in one folder with filename_employeenumber as shown below .
/server/user/home/xml/allemployees/Employeedetails-e1100123.xml
/server/user/home/xml/allemployees/Employeesalary-e1100123.xml
/server/user/home/xml/allemployees/Employeeleaves-e1100123.xml
/server/user/home/xml/allemployees/Employeestatus-e1100123.xml
/server/user/home/xml/allemployees/Employeedetails-e1100155.xml
/server/user/home/xml/allemployees/Employeesalary-e1100155.xml
/server/user/home/xml/allemployees/Employeeleaves-e1100155.xml
/server/user/home/xml/allemployees/Employeestatus-e1100155.xml
How to write a code in unix shell script ?
Thank you ......
Sai
Bare bone of your desired script could be like so:
script.sh
#!/bin/bash
base_dir=/server/user/home/xml
all_dir=${base_dir}/allemployes
e_dirs=$(ls -d ${base_dir}/e*)
e_files=(Employeedetails.xml Employeesalary.xml Employeestatus.xml Employeeleaves.xml)
if [ ! -d "${base_dir}" ]; then
echo "ERROR: ${base_dir} doesn't exists!"
fi
if [ ! -d "${all_dir}" ]; then
echo "INFO: Creating all employees directory at ${all_dir}"
mkdir "${all_dir}"
fi
for e in ${e_dirs}; do
pushd "${e}" > /dev/null
for f in ${e_files[#]}; do
if [ ! -f ${f} ]; then
echo "ERROR: Missing ${f} file for employee $(basename ${e})"
else
new_file=${all_dir}/${f%.*}-$(basename ${e}).xml
cp "${e}/${f}" "${new_file}"
fi
done
popd > /dev/null
done
Notes:
Script assumes that in xml/ folder are only employees folder and every folder starts with lower character e, hence we do e_dirs=$(ls -d ${base_dir}/e*)
Script works only with files Employeedetails.xml Employeesalary.xml Employeestatus.xml Employeeleaves.xml file, ignoring others, in each employee folder.
Script creates allemployes/ folder if it does not exists.
Script makes copies instead of moving (not destructive), so you can use mv instead of cp
I haven't tested the script! So please do not run it on live data :). Feel free to modify it.
Use this script :
#!/bin/bash
sourceDir=$1
destDir=$2
dirname=$(basename $1)
if [ ! -d $2 ] ; then
mkdir $2
fi
for I in $sourceDir/*
do
fname=$(basename $I)
prefix=$(echo $fname | awk -F "." 'NF{NF-=1};1')
suffix=$(echo $fname | awk -F "." '{print $NF}')
target="$destDir/$prefix-$dirname.$suffix"
cp $I $target
# I use cp(Copy) command , you can replace this to mv(Move) command .
echo $target
done
Usage :
./script.sh [SOURCEDIR] [DESTDIR]
Example :
./script.sh /server/user/home/xml/e1100123/ /server/user/home/xml/allemployees/

how to call a command from a bash script in Linux

I am trying to call a command from a script (defined as functions inside .bashrc) as following
the first function hooks the rm command by deleting any file except a file named "forensicfile1"
#hooking rm command
function rm(){
if [[ $1 = "forensicfile1" ]]; then
echo "$1 file is deleted by user ">forensicaudit
else
rm $1
fi
}
the second hooking function is trying to hook the ls command by listing all files except the forensicfile1
#hooking ls command
function ls(){
if [[ $PWD = "/home/waleed/forensicdata" ]]
then
alias myls='ls "/home/waleed/forensicdata" -1 "forensicfile1" '
myls
else
ls $PWD
fi
}
however the following lines are not working
rm $1 // within the first function
alias myls='ls "/home/waleed/forensicdata" -1 "forensicfile1" '
myls
ls $PWD
can anyone help?

ksh --> is it possible to run a command on the parent shell (main shell) from within a subshell?

#!/usr/bin/ksh
if [ $# -ne 1 ]; then
echo "[*]\t Please see usage..."
echo "[*]\t Usage: $0 <store_number>"
exit 1
fi
if [ -z "$1" ]; then
echo "[*]\t Please see usage..."
echo "[*]\t Usage: $0 <store_number>"
exit 1
fi
Store_Number=$1
EPS_Directory="/apps/epsadmin_90000"$Store_Number"/EPS"
cd $EPS_Directory
I am trying to write a simple script that will change my directory in my main shell.
I have it working to change directory within the sub-shell (shown above), but obviously when the script is done running it kicks me back out to the outer shell and I am back in my original directory.
Is it possible to pass a command to the outer shell, from within a sub-shell? Can I pass a cd command to the outer shell?
For example if I run:
./cd.sh 2001
I would like my directory to be:
/apps/epsadmin_900002001/EPS
Once I return to the outer shell.
No, this is not possible.
Instead, you can make a function:
mycd() {
if [ $# -ne 1 ]; then
echo "[*]\t Please see usage..."
echo "[*]\t Usage: $0 <store_number>"
return 1
fi
if [ -z "$1" ]; then
echo "[*]\t Please see usage..."
echo "[*]\t Usage: $0 <store_number>"
return 1
fi
Store_Number=$1
EPS_Directory="/apps/epsadmin_90000$Store_Number/EPS"
cd "$EPS_Directory"
}
... and store it in a file of its own and source it:
. $HOME/.fun/mycd.sh
Shell functions run in the main process, unlike scripts which run in subprocesses.
Thanks for all your help! This is my solution.
# create dj file in /users/(YOUR_NUID) directory
# paste the dj function into this file. (vi dj) (hit i to enter edit mode) (right click to paste) (hit esc) (type :wq)
# source the dj file containing dj() functon by adding this to .profile:
# . $HOME/dj
# reload .profile by typing . ./.profile
# then to run the function simply type dj <storenumber> to jump between EPS directory folders.
dj(){
Store_Number=$1
EPS_Directory="/apps/epsadmin_90000"$Store_Number"/EPS"
if [ -e $(echo $EPS_Directory) ]; then
cd $EPS_Directory
echo "You are now in directory: $EPS_Directory"
else
echo "Directory $EPS_Directory does not exist."
fi
}

How do I check if a directory exists or not in a Bash shell script?

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]

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