Bash script with argument that makes file executable - linux

I need to make a bash script that checks if the file or directory exists,then if the file does,it checks the executable permission.I need to modify the script to be able to give a file executable permissions from an argument.
Example: Console input ./exist.sh +x file_name should make the file executable.
This is the unfinished code that checks if the file/directory exists and if the file is executable or not. I need to add the chmod argument part.
#!/bin/bash
file=$1
if [ -x $file ]; then
echo "The file '$file' exists and it is exxecutable"
else
echo "The file '$file' is not executable (or does not exist)"
fi
if [ -d $file ]; then
echo "There is a directory named '$file'"
else
echo "There is no directory named '$file'"
fi

If you have optional arguments to your script, you need to check for them first.
In the case of just a couple of simple arguments, it would be simpler to check for them explicitly.
MAKEEXECUTABLE=0
while [ "${1:0:1}" = "+" ]; do
case $1 in
"+x")
MAKEEXECUTABLE=1
shift
;;
*)
echo "Unknown option '$1'"
exit
esac
done
file=$1
Then after you have determined that the file is not executable
if [ $MAKEEXECUTABLE -eq 1 ]; then
chmod +x $file
fi
Should you decide to add more complex options, you may want to use something like getops:example of how to use getopts in bash

Add chmod something like:
if [ ! -x "$file" ]; then
chmod +x $file
fi
This means if file does not have execute persmission, then add execute permission for the user.

Related

Using ls command result in a loop

I want to use the result of ls command in a loop to check if for example the first line is a directory, second etc.
For example I have this folder that contains one directory the script should display:
18_05_2018 is directory
enter image description here
Create a file named is_file_or_directory.sh containing:
cd "$1" || echo "Please specify a path" && exit
for i in *; do
if [[ -d $i ]]; then
echo "$i is a directory"
elif [[ -f $i ]]; then
echo "$i is a file"
else
echo "$i is not valid"
exit 1
fi
done
Make that file executable with:
sudo chmod +x is_file_or_directory.sh
Run the script specifying as a parameter the path that you want to analyze:
./is_file_or_directory.sh /root/scripts/
Output:
jeeves ~/scripts/stack # ./is_file_or_dir.sh /root/scripts/
databe.py is a file
is_file_or_dir.sh is a file
mysql_flask.py is a file
test is a directory
Here's a more detailed explanation of what is happening under the hood. The variable $1 is, in Bash, the first parameter sent to the script. In our case it is the path where the script will perform its actions. Then we use the variable $i in the loop.
$i content will be every file / folder name in the path $1. With -d and -f we check if $i is a file or a folder.

check to see if filepath exists bash linux script [duplicate]

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]

How to write a linux script that copies all the files and directories under one directory to another directory?

I am having a bit of a problem with a script I am trying to create.
This script should take two arguments which are the source directory and the destination directory and if the user enters less than 2 arguments it should print out an error message and exit. In addition this script should check if the source directory exists or not, if not it should print out an error message and exit. Also the script should check if the destination directory exists or not, if not it should create that directory. Finally the script should copy the files from the source directory to the destination directory.
This is my attempt so far:
if (( $# < 2 ));
echo "Error: Too few arguments supplied"
exit 1
if [ -d "src_dir" ]
then
echo "Directory src_dir exists."
else
echo "Error: Directory src_dir does not exist."
fi
if [ -d "dst_dir" ]
then
echo "Directory dst_dir exists."
else
mkdir dst_dir
cp -r src_dir/* dst_dir
fi
Any help would be really appreciated. Thanks in advance!
For checking the correct number of arguments: Check number of arguments passed to a Bash script
if [ "$#" -ne 2 ]; then
echo "Error: Too few arguments supplied"
exit 1
fi
For checking if the directory exists or not: Check if a directory exists in a shell script
if [ ! -d "$1" ]; then
echo "Error: Directory $1 does not exist."
exit 1
fi
For making the dir on second argument: How to use Bash to create a folder if it doesn't already exist?
mkdir -p $2
Finally, just copy them all:
cp -r $1/* $2/

linux zip and exclude dir via bash/shell script

I am trying to write a bash/shell script to zip up a specific folder and ignore certain sub-dirs in that folder.
This is the folder I am trying to zip "sync_test5":
My bash script generates an ignore list (based on) and calls the zip function like this:
#!/bin/bash
SYNC_WEB_ROOT_BASE_DIR="/home/www-data/public_html"
SYNC_WEB_ROOT_BACKUP_DIR="sync_test5"
SYNC_WEB_ROOT_IGNORE_DIR="dir_to_ignore dir2_to_ignore"
ignorelist=""
if [ "$SYNC_WEB_ROOT_IGNORE_DIR" != "" ];
then
for ignoredir in $SYNC_WEB_ROOT_IGNORE_DIR
do
ignorelist="$ignorelist $SYNC_WEB_ROOT_BACKUP_DIR/$ignoredir/**\*"
done
fi
FILE="$SYNC_BACKUP_DIR/$DATETIMENOW.website.zip"
cd $SYNC_WEB_ROOT_BASE_DIR;
zip -r $FILE $SYNC_WEB_ROOT_BACKUP_DIR -x $ignorelist >/dev/null
echo "Done"
Now this script runs without error, however it is not ignoring/excluding the dirs I've specified.
So, I had the shell script output the command it tried to run, which was:
zip -r 12-08-2014_072810.website.zip sync_test5 -x sync_test5/dir_to_ignore/**\* sync_test5/dir2_to_ignore/**\*
Now If I run the above command directly in putty like this, it works:
So, why doesn't my shell script exclude working as intended? the command that is being executed is identical (in shell and putty directly).
Because backslash quotings in a variable after word splitting are not evaluated.
If you have a='123\4', echo $a would give
123\4
But if you do it directly like echo 123\4, you'd get
1234
Clearly the arguments you pass with the variable and without the variables are different.
You probably just meant to not quote your argument with backslash:
ignorelist="$ignorelist $SYNC_WEB_ROOT_BACKUP_DIR/$ignoredir/***"
Btw, what actual works is a non-evaluated glob pattern:
zip -r 12-08-2014_072810.website.zip sync_test5 -x 'sync_test5/dir_to_ignore/***' 'sync_test5/dir2_to_ignore/***'
You can verify this with
echo zip -r 12-08-2014_072810.website.zip sync_test5 -x sync_test5/dir_to_ignore/**\* sync_test5/dir2_to_ignore/**\*
And this is my suggestion:
#!/bin/bash
SYNC_WEB_ROOT_BASE_DIR="/home/www-data/public_html"
SYNC_WEB_ROOT_BACKUP_DIR="sync_test5"
SYNC_WEB_ROOT_IGNORE_DIR=("dir_to_ignore" "dir2_to_ignore")
IGNORE_LIST=()
if [[ -n $SYNC_WEB_ROOT_IGNORE_DIR ]]; then
for IGNORE_DIR in "${SYNC_WEB_ROOT_IGNORE_DIR[#]}"; do
IGNORE_LIST+=("$SYNC_WEB_ROOT_BACKUP_DIR/$IGNORE_DIR/***") ## "$SYNC_WEB_ROOT_BACKUP_DIR/$IGNORE_DIR/*" perhaps is enough?
done
fi
FILE="$SYNC_BACKUP_DIR/$DATETIMENOW.website.zip" ## Where is $SYNC_BACKUP_DIR set?
cd "$SYNC_WEB_ROOT_BASE_DIR";
zip -r "$FILE" "$SYNC_WEB_ROOT_BACKUP_DIR" -x "${IGNORE_LIST[#]}" >/dev/null
echo "Done"
This is what I ended up with:
#!/bin/bash
# This script zips a directory, excluding specified files, types and subdirectories.
# while zipping the directory it excludes hidden directories and certain file types
[[ "`/usr/bin/tty`" == "not a tty" ]] && . ~/.bash_profile
DIRECTORY=$(cd `dirname $0` && pwd)
if [[ -z $1 ]]; then
echo "Usage: managed_directory_compressor /your-directory/ zip-file-name"
else
DIRECTORY_TO_COMPRESS=${1%/}
ZIPPED_FILE="$2.zip"
COMPRESS_IGNORE_FILE=("\.git" "*.zip" "*.csv" "*.json" "gulpfile.js" "*.rb" "*.bak" "*.swp" "*.back" "*.merge" "*.txt" "*.sh" "bower_components" "node_modules")
COMPRESS_IGNORE_DIR=("bower_components" "node_modules")
IGNORE_LIST=("*/\.*" "\.* "\/\.*"")
if [[ -n $COMPRESS_IGNORE_FILE ]]; then
for IGNORE_FILES in "${COMPRESS_IGNORE_FILE[#]}"; do
IGNORE_LIST+=("$DIRECTORY_TO_COMPRESS/$IGNORE_FILES/*")
done
for IGNORE_DIR in "${COMPRESS_IGNORE_DIR[#]}"; do
IGNORE_LIST+=("$DIRECTORY_TO_COMPRESS/$IGNORE_DIR/")
done
fi
zip -r "$ZIPPED_FILE" "$DIRECTORY_TO_COMPRESS" -x "${IGNORE_LIST[#]}" # >/dev/null
# echo zip -r "$ZIPPED_FILE" "$DIRECTORY_TO_COMPRESS" -x "${IGNORE_LIST[#]}" # >/dev/null
echo $DIRECTORY_TO_COMPRESS "compressed as" $ZIPPED_FILE.
fi
After a few trial and error, I have managed to fix this problem by changing this line:
ignorelist="$ignorelist $SYNC_WEB_ROOT_BACKUP_DIR/$ignoredir/**\*"
to:
ignorelist="$ignorelist $SYNC_WEB_ROOT_BACKUP_DIR/$ignoredir/***"
Not sure why this worked, but it does :)

Bash script function call error

I am writing my first Bash script and am running into a syntax issue with a function call.
Specifically, I want to invoke my script like so:
sh myscript.sh -d=<abc>
Where <abc> is the name of a specific directory inside of a fixed parent directory (~/app/dropzone). If the child <abc> directory doesn't exist, I want the script to create it before going to that directory. If the user doesn't invoke the script with a -d argument at all, I want the script to exist with a simple usage message. Here's my best attempt at the script so far:
#!/bin/bash
dropzone="~/app/dropzone"
# If the directory the script user specified exists, overwrite dropzone value with full path
# to directory. If the directory doesn't exist, first create it. If user failed to specify
# -d=<someDirName>, exit the script with a usage statement.
validate_args() {
args=$(getopt d: "$*")
set -- $args
dir=$2
if [ "$dir" ]
then
if [ ! -d "${dropzone}/targets/$dir" ]
then
mkdir ${dropzone}/targets/$dir
fi
dropzone=${dropzone}/targets/$dir
else
usage
fi
}
usage() {
echo "Usage: $0" >&2
exit 1
}
# Validate script arguments.
validate_args $1
# Go to the dropzone directory.
cd dropzone
echo "Arrived at dropzone $dropzone."
# The script will now do other stuff, now that we're in the "dropzone".
# ...etc.
When I try running this I get the following error:
myUser#myMachine:~/app/scripts$ sh myscript.sh -dyoyo
mkdir: cannot create directory `/home/myUser/app/dropzone/targets/yoyo': No such file or directory
myscript.sh: 33: cd: can't cd to dropzone
Arrived at dropzone /home/myUser/app/dropzone/targets/yoyo.
Where am I going wrong, and is my general approach even correct? Thanks in advance!
Move the function definitions to the top of the script (below the hash-bang). bash is objecting to the undefined (at that point) call to validate_args. usage definition should precede the definition of validate_args.
There also should be spacing in the if tests "[ " and " ]".
if [ -d "$dropzone/targets/$1" ]
The getopt test for option d should be-:
if [ "$(getopt d "$1")" ]
Here is a version of validate_args that works for me.
I also had to change the drop zone as on my shell ~ wouldn't expand in mkdir command.
dropzone="/home/suspectus/app/dropzone"
validate_args() {
args=$(getopt d: "$*")
set -- $args
dir=$2
if [ "$dir" ]
then
if [ ! -d "${dropzone}/targets/$dir" ]
then
mkdir ${dropzone}/targets/$dir
fi
dropzone=${dropzone}/targets/$dir
else
usage
fi
}
To pass in all args use $* as parameter -:
validate_args $*
And finally call the script like this for getopt to parse correctly-:
myscript.sh -d dir_name
When invoked, a function is indistinguishable from a command — so you don't use parentheses:
validate_args($1) # Wrong
validate_args $1 # Right
Additionally, as suspectus points out in his answer, functions must be defined before they are invoked. You can see this with the script:
usage
usage()
{
echo "Usage: $0" >&2
exit 1
}
which will report usage: command not found assuming you don't have a command or function called usage available. Place the invocation after the function definition and it will work fine.
Your chosen interface is not the standard Unix calling convention for commands. You'd normally use:
dropzone -d subdir
rather than
dropzone -d=subdir
However, we can handle your chosen interface (but not using getopts, the built-in command interpreter, and maybe not using GNU getopt either, and certainly not using getopt as you tried to do so). Here's workable code supporting -d=subdir:
#!/bin/bash
dropzone="$HOME/app/dropzone/targets"
validate_args()
{
case "$1" in
(-d=*) dropzone="$dropzone/${1#-d=}"; mkdir -p $dropzone;;
(*) usage;;
esac
}
usage()
{
echo "Usage: $0 -d=dropzone" >&2
exit 1
}
# Validate script arguments.
validate_args $1
# Go to the dropzone directory.
cd $dropzone || exit 1
echo "Arrived at dropzone $dropzone."
# The script will now do other stuff, now that we're in the "dropzone".
# ...etc.
Note the cautious approach with the cd $dropzone || exit 1; if the cd fails, you definitely do not want to continue in the wrong directory.
Using the getopts built-in command interpreter:
#!/bin/bash
dropzone="$HOME/app/dropzone/targets"
usage()
{
echo "Usage: $0 -d dropzone" >&2
exit 1
}
while getopts d: opt
do
case "$opt" in
(d) dropzone="$dropzone/$OPTARG"; mkdir -p $dropzone;;
(*) usage;;
esac
done
shift $(($OPTIND - 1))
# Go to the dropzone directory.
cd $dropzone || exit 1
echo "Arrived at dropzone $dropzone."
# The script will now do other stuff, now that we're in the "dropzone".
# ...etc.

Resources