I would like to use custom identity file when using rsync, but only if the file exists, otherwise I don't want to bother with custom ssh command for rsync. I am having problems with quotes. See examples.
Desired command if identity file exists
rsync -e "ssh -i '/tmp/id_rsa'" /tmp/dir/ u#h:/tmp/dir
Desired command if identity file does not exist
rsync /tmp/dir/ u#h:/tmp/dir
I wanted to create a variable that would contain -e "ssh -i '/tmp/id_rsa'" and use it as follows
rsync ${identityArg} /tmp/dir/ u#h:/tmp/dir
This variable would be either empty or contain desired ssh command.
An example way I fill the variable (I have tried many ways)
IDENTITY_FILE="/tmp/id_rsa"
if [ -f "${IDENTITY_FILE}" ]; then
identityArg="-e 'ssh -i \"${IDENTITY_FILE}\"'"
fi
The problem is that quotes are always wrong in the command and I end up with commands similar to these ones (set -x is set in the script and this is the output)
rsync -e '\ssh' -i '"/tmp/id_rsa"'\''' /tmp/dir/ u#h:/tmp/dir
There is something I do not get about quotation in bash. If you have any good resource about usage of single and double quotes in bash script I would like to read it.
You want to add two positional parameters: -e and ssh -i '/tmp/id_rsa', where /tmp/id_rsa is an expanded variable. You should use an array for this:
args=(/tmp/dir/ u#h:/tmp/dir)
idfile=/tmp/id_rsa
# Let [[ ... ]] do the quoting
if [[ -f $idfile ]]; then
# Prepend two parameters to args array
args=(-e "ssh -i '$idfile'" "${args[#]}")
fi
rsync "${args[#]}"
I'm not convinced the inner single quotes are necessary for ssh -i, but this expands to exactly the commands shown in the question.
Try like this
id=/tmp/id_rsa
[[ -e $id ]] && o1='-e' o2="ssh -i '$id'"
echo $o1 "$o2" /tmp/dir/ u#h:/tmp/dir
Trying to properly escape quotes is tricky. Better to try to leverage existing constructs. Few alternatives, depending on the situations
If the name of the identity file only contain simple characters (no spaces, wildcard, etc.) consider not wrapping it in quotes. In this case, you can
IDENTITY_FILE="/tmp/id_rsa"
if [ -f "${IDENTITY_FILE}" ]; then
identityArg="-e 'ssh -i ${IDENTITY_FILE}'"
fi
...
rsync $identityArg ...
Another option is to always pass in the command (ssh or 'ssh -I ...'). This will automatically take care for special characters in the identity file.
IDENTITY_FILE="/tmp/id_rsa"
if [ -f "${IDENTITY_FILE}" ]; then
identityArg="-i '${IDENTITY_FILE}'"
fi
rsync -e "ssh $identityArg" ...
Third alternative is to use array to create the arguments to rsync, and let the shell escape the characters as needed. This will allow any character in the identity file.
IDENTITY_FILE="/tmp/id_rsa"
if [ -f "${IDENTITY_FILE}" ]; then
identityArg=(-e "ssh -i '${IDENTITY_FILE}'")
fi
rsync "${identityArg[#]}" ...
I am trying to upload multiple files from one folder to a ftp site and wrote this script:
#!/bin/bash
for i in '/dir/*'
do
if [-f /dir/$i]; then
HOST='x.x.x.x'
USER='username'
PASSWD='password'
DIR=archives
File=$i
ftp -n $HOST << END_SCRIPT
quote USER $USER
quote PASS $PASSWD
ascii
put $FILE
quit
END_SCRIPT
fi
It is giving me following error when I try to execute:
username#host:~/Documents/Python$ ./script.sh
./script.sh: line 22: syntax error: unexpected end of file
I can't seem to get this to work. Any help is much appreciated.
Thanks,
Mayank
It's complaining because your for loop does not have a done marker to indicate the end of the loop. You also need more spaces in your if:
if [ -f "$i" ]; then
Recall that [ is actually a command, and it won't be recognized if it doesn't appear as such.
And... if you single quote your glob (at the for) like that, it won't be expanded. No quotes there, but double quotes when using $i. You probably also don't want to include the /dir/ part when you use $i as it's included in your glob.
If I'm not mistaken, ncftp can take wildcard arguments:
ncftpput -u username -p password x.x.x.x archives /dir/*
If you don't already have it installed, it's likely available in the standard repo for your OS.
First, the literal, fixing-your-script answer:
#!/bin/bash
# no reason to set variables that don't change inside the loop
host='x.x.x.x'
user='username'
password='password'
dir=archives
for i in /dir/*; do # no quotes if you want the wildcard to be expanded!
if [ -f "$i" ]; then # need double quotes and whitespace here!
file=$i
ftp -n "$host" <<END_SCRIPT
quote USER $user
quote PASS $password
ascii
put $file $dir/$file
quit
END_SCRIPT
fi
done
Next, the easy way:
lftp -e 'mput -a *.i' -u "$user,$password" "ftp://$host/"
(yes, lftp expands the wildcard internally, rather than expecting this to be done by the outer shell).
First of all my apologies in not making myself clear in the question. My actual task was to copy a file from local folder to a SFTP site and then move the file to an archive folder. Since the SFTP is hosted by a vendor I cannot use the key sharing (vendor limitation. Also, SCP will require password entering if used in a shell script so I have to use SSHPASS. SSHPASS is in the Ubuntu repo however for CentOS it needs to be installed from here
Current thread and How to run the sftp command with a password from Bash script? did gave me better understanding on how to write the script and I will share my solution here:
#!/bin/bash
#!/usr/bin
for i in /dir/*; do
if [ -f "$i" ]; then
file=$i
export SSHPASS=password
sshpass -e sftp -oBatchMode=no -b - user#ftp.com << !
cd foldername/foldername
put $file
bye
!
mv $file /somedir/test
fi
done
Thanks everyone for all the responses!
--Mayank
I'm trying to write a bash script that "wraps" whatever the user wants to invoke (and its parameters) sourcing a fixed file just before actually invoking it.
To clarify: I have a "ConfigureMyEnvironment.bash" script that must be sourced before starting certain executables, so I'd like to have a "LaunchInMyEnvironment.bash" script that you can use as in:
LaunchInMyEnvironment <whatever_executable_i_want_to_wrap> arg0 arg1 arg2
I tried the following LaunchInMyEnvironment.bash:
#!/usr/bin/bash
launchee="$#"
if [ -e ConfigureMyEnvironment.bash ];
then source ConfigureMyEnvironment.bash;
fi
exec "$launchee"
where I have to use the "launchee" variable to save the $# var because after executing source, $# becomes empty.
Anyway, this doesn't work and fails as follows:
myhost $ LaunchInMyEnvironment my_executable -h
myhost $ /home/me/LaunchInMyEnvironment.bash: line 7: /home/bin/my_executable -h: No such file or directory
myhost $ /home/me/LaunchInMyEnvironment.bash: line 7: exec: /home/bin/my_executable -h: cannot execute: No such file or directory
That is, it seems like the "-h" parameter is being seen as part of the executable filename and not as a parameter... But it doesn't really make sense to me.
I tried also to use $* instead of $#, but with no better outcoume.
What I'm doing wrong?
Andrea.
Have you tried to remove double quotes in exec command?
Try this:
#!/usr/bin/bash
typeset -a launchee
launchee=("$#")
if [ -e ConfigureMyEnvironment.bash ];
then source ConfigureMyEnvironment.bash;
fi
exec "${launchee[#]}"
That will use arrays for storing arguments, so it will handle even calls like "space delimited string" and "string with ; inside"
Upd: simple example
test_array() { abc=("$#"); for x in "${abc[#]}"; do echo ">>$x<<"; done; }
test_array "abc def" ghi
should give
>>abc def<<
>>ghi<<
You might want to try this (untested):
#!/usr/bin/bash
launchee="$1"
shift
if [ -e ConfigureMyEnvironment.bash ];
then source ConfigureMyEnvironment.bash;
fi
exec "$launchee" $#
The syntax for exec is exec command [arguments], however becuase you've quoted $launchee, this is treated as a single argument - i.e., the command, rather than a command and it's arguments. Another variation may be to simply do: exec $#
Just execute it normally without exec
#!/usr/bin/bash
launchee="$#"
if [ -e ConfigureMyEnvironment.bash ];
then source ConfigureMyEnvironment.bash;
fi
$launchee
Try dividing your list of argumets:
ALL_ARG="${#}"
Executable="${1}"
Rest_of_Args=${ALL_ARG##$Executable}
And try then:
$Executable $Rest_of_Args
(or exec $Executable $Rest_of_Args)
Debugger
How would I validate that a program exists, in a way that will either return an error and exit, or continue with the script?
It seems like it should be easy, but it's been stumping me.
Answer
POSIX compatible:
command -v <the_command>
Example use:
if ! command -v <the_command> &> /dev/null
then
echo "<the_command> could not be found"
exit
fi
For Bash specific environments:
hash <the_command> # For regular commands. Or...
type <the_command> # To check built-ins and keywords
Explanation
Avoid which. Not only is it an external process you're launching for doing very little (meaning builtins like hash, type or command are way cheaper), you can also rely on the builtins to actually do what you want, while the effects of external commands can easily vary from system to system.
Why care?
Many operating systems have a which that doesn't even set an exit status, meaning the if which foo won't even work there and will always report that foo exists, even if it doesn't (note that some POSIX shells appear to do this for hash too).
Many operating systems make which do custom and evil stuff like change the output or even hook into the package manager.
So, don't use which. Instead use one of these:
command -v foo >/dev/null 2>&1 || { echo >&2 "I require foo but it's not installed. Aborting."; exit 1; }
type foo >/dev/null 2>&1 || { echo >&2 "I require foo but it's not installed. Aborting."; exit 1; }
hash foo 2>/dev/null || { echo >&2 "I require foo but it's not installed. Aborting."; exit 1; }
(Minor side-note: some will suggest 2>&- is the same 2>/dev/null but shorter – this is untrue. 2>&- closes FD 2 which causes an error in the program when it tries to write to stderr, which is very different from successfully writing to it and discarding the output (and dangerous!))
If your hash bang is /bin/sh then you should care about what POSIX says. type and hash's exit codes aren't terribly well defined by POSIX, and hash is seen to exit successfully when the command doesn't exist (haven't seen this with type yet). command's exit status is well defined by POSIX, so that one is probably the safest to use.
If your script uses bash though, POSIX rules don't really matter anymore and both type and hash become perfectly safe to use. type now has a -P to search just the PATH and hash has the side-effect that the command's location will be hashed (for faster lookup next time you use it), which is usually a good thing since you probably check for its existence in order to actually use it.
As a simple example, here's a function that runs gdate if it exists, otherwise date:
gnudate() {
if hash gdate 2>/dev/null; then
gdate "$#"
else
date "$#"
fi
}
Alternative with a complete feature set
You can use scripts-common to reach your need.
To check if something is installed, you can do:
checkBin <the_command> || errorMessage "This tool requires <the_command>. Install it please, and then run this tool again."
The following is a portable way to check whether a command exists in $PATH and is executable:
[ -x "$(command -v foo)" ]
Example:
if ! [ -x "$(command -v git)" ]; then
echo 'Error: git is not installed.' >&2
exit 1
fi
The executable check is needed because bash returns a non-executable file if no executable file with that name is found in $PATH.
Also note that if a non-executable file with the same name as the executable exists earlier in $PATH, dash returns the former, even though the latter would be executed. This is a bug and is in violation of the POSIX standard. [Bug report] [Standard]
Edit: This seems to be fixed as of dash 0.5.11 (Debian 11).
In addition, this will fail if the command you are looking for has been defined as an alias.
I agree with lhunath to discourage use of which, and his solution is perfectly valid for Bash users. However, to be more portable, command -v shall be used instead:
$ command -v foo >/dev/null 2>&1 || { echo "I require foo but it's not installed. Aborting." >&2; exit 1; }
Command command is POSIX compliant. See here for its specification: command - execute a simple command
Note: type is POSIX compliant, but type -P is not.
It depends on whether you want to know whether it exists in one of the directories in the $PATH variable or whether you know the absolute location of it. If you want to know if it is in the $PATH variable, use
if which programname >/dev/null; then
echo exists
else
echo does not exist
fi
otherwise use
if [ -x /path/to/programname ]; then
echo exists
else
echo does not exist
fi
The redirection to /dev/null/ in the first example suppresses the output of the which program.
I have a function defined in my .bashrc that makes this easier.
command_exists () {
type "$1" &> /dev/null ;
}
Here's an example of how it's used (from my .bash_profile.)
if command_exists mvim ; then
export VISUAL="mvim --nofork"
fi
Expanding on #lhunath's and #GregV's answers, here's the code for the people who want to easily put that check inside an if statement:
exists()
{
command -v "$1" >/dev/null 2>&1
}
Here's how to use it:
if exists bash; then
echo 'Bash exists!'
else
echo 'Your system does not have Bash'
fi
Try using:
test -x filename
or
[ -x filename ]
From the Bash manpage under Conditional Expressions:
-x file
True if file exists and is executable.
To use hash, as #lhunath suggests, in a Bash script:
hash foo &> /dev/null
if [ $? -eq 1 ]; then
echo >&2 "foo not found."
fi
This script runs hash and then checks if the exit code of the most recent command, the value stored in $?, is equal to 1. If hash doesn't find foo, the exit code will be 1. If foo is present, the exit code will be 0.
&> /dev/null redirects standard error and standard output from hash so that it doesn't appear onscreen and echo >&2 writes the message to standard error.
Command -v works fine if the POSIX_BUILTINS option is set for the <command> to test for, but it can fail if not. (It has worked for me for years, but I recently ran into one where it didn't work.)
I find the following to be more failproof:
test -x "$(which <command>)"
Since it tests for three things: path, existence and execution permission.
There are a ton of options here, but I was surprised no quick one-liners. This is what I used at the beginning of my scripts:
[[ "$(command -v mvn)" ]] || { echo "mvn is not installed" 1>&2 ; exit 1; }
[[ "$(command -v java)" ]] || { echo "java is not installed" 1>&2 ; exit 1; }
This is based on the selected answer here and another source.
If you check for program existence, you are probably going to run it later anyway. Why not try to run it in the first place?
if foo --version >/dev/null 2>&1; then
echo Found
else
echo Not found
fi
It's a more trustworthy check that the program runs than merely looking at PATH directories and file permissions.
Plus you can get some useful result from your program, such as its version.
Of course the drawbacks are that some programs can be heavy to start and some don't have a --version option to immediately (and successfully) exit.
Check for multiple dependencies and inform status to end users
for cmd in latex pandoc; do
printf '%-10s' "$cmd"
if hash "$cmd" 2>/dev/null; then
echo OK
else
echo missing
fi
done
Sample output:
latex OK
pandoc missing
Adjust the 10 to the maximum command length. It is not automatic, because I don't see a non-verbose POSIX way to do it:
How can I align the columns of a space separated table in Bash?
Check if some apt packages are installed with dpkg -s and install them otherwise.
See: Check if an apt-get package is installed and then install it if it's not on Linux
It was previously mentioned at: How can I check if a program exists from a Bash script?
I never did get the previous answers to work on the box I have access to. For one, type has been installed (doing what more does). So the builtin directive is needed. This command works for me:
if [ `builtin type -p vim` ]; then echo "TRUE"; else echo "FALSE"; fi
I wanted the same question answered but to run within a Makefile.
install:
#if [[ ! -x "$(shell command -v ghead)" ]]; then \
echo 'ghead does not exist. Please install it.'; \
exit -1; \
fi
It could be simpler, just:
#!/usr/bin/env bash
set -x
# if local program 'foo' returns 1 (doesn't exist) then...
if ! type -P foo; then
echo 'crap, no foo'
else
echo 'sweet, we have foo!'
fi
Change foo to vi to get the other condition to fire.
hash foo 2>/dev/null: works with Z shell (Zsh), Bash, Dash and ash.
type -p foo: it appears to work with Z shell, Bash and ash (BusyBox), but not Dash (it interprets -p as an argument).
command -v foo: works with Z shell, Bash, Dash, but not ash (BusyBox) (-ash: command: not found).
Also note that builtin is not available with ash and Dash.
zsh only, but very useful for zsh scripting (e.g. when writing completion scripts):
The zsh/parameter module gives access to, among other things, the internal commands hash table. From man zshmodules:
THE ZSH/PARAMETER MODULE
The zsh/parameter module gives access to some of the internal hash ta‐
bles used by the shell by defining some special parameters.
[...]
commands
This array gives access to the command hash table. The keys are
the names of external commands, the values are the pathnames of
the files that would be executed when the command would be in‐
voked. Setting a key in this array defines a new entry in this
table in the same way as with the hash builtin. Unsetting a key
as in `unset "commands[foo]"' removes the entry for the given
key from the command hash table.
Although it is a loadable module, it seems to be loaded by default, as long as zsh is not used with --emulate.
example:
martin#martin ~ % echo $commands[zsh]
/usr/bin/zsh
To quickly check whether a certain command is available, just check if the key exists in the hash:
if (( ${+commands[zsh]} ))
then
echo "zsh is available"
fi
Note though that the hash will contain any files in $PATH folders, regardless of whether they are executable or not. To be absolutely sure, you have to spend a stat call on that:
if (( ${+commands[zsh]} )) && [[ -x $commands[zsh] ]]
then
echo "zsh is available"
fi
The which command might be useful. man which
It returns 0 if the executable is found and returns 1 if it's not found or not executable:
NAME
which - locate a command
SYNOPSIS
which [-a] filename ...
DESCRIPTION
which returns the pathnames of the files which would
be executed in the current environment, had its
arguments been given as commands in a strictly
POSIX-conformant shell. It does this by searching
the PATH for executable files matching the names
of the arguments.
OPTIONS
-a print all matching pathnames of each argument
EXIT STATUS
0 if all specified commands are
found and executable
1 if one or more specified commands is nonexistent
or not executable
2 if an invalid option is specified
The nice thing about which is that it figures out if the executable is available in the environment that which is run in - it saves a few problems...
Use Bash builtins if you can:
which programname
...
type -P programname
For those interested, none of the methodologies in previous answers work if you wish to detect an installed library. I imagine you are left either with physically checking the path (potentially for header files and such), or something like this (if you are on a Debian-based distribution):
dpkg --status libdb-dev | grep -q not-installed
if [ $? -eq 0 ]; then
apt-get install libdb-dev
fi
As you can see from the above, a "0" answer from the query means the package is not installed. This is a function of "grep" - a "0" means a match was found, a "1" means no match was found.
This will tell according to the location if the program exist or not:
if [ -x /usr/bin/yum ]; then
echo "This is Centos"
fi
I'd say there isn't any portable and 100% reliable way due to dangling aliases. For example:
alias john='ls --color'
alias paul='george -F'
alias george='ls -h'
alias ringo=/
Of course, only the last one is problematic (no offence to Ringo!). But all of them are valid aliases from the point of view of command -v.
In order to reject dangling ones like ringo, we have to parse the output of the shell built-in alias command and recurse into them (command -v isn't a superior to alias here.) There isn't any portable solution for it, and even a Bash-specific solution is rather tedious.
Note that a solution like this will unconditionally reject alias ls='ls -F':
test() { command -v $1 | grep -qv alias }
If you guys/gals can't get the things in answers here to work and are pulling hair out of your back, try to run the same command using bash -c. Just look at this somnambular delirium. This is what really happening when you run $(sub-command):
First. It can give you completely different output.
$ command -v ls
alias ls='ls --color=auto'
$ bash -c "command -v ls"
/bin/ls
Second. It can give you no output at all.
$ command -v nvm
nvm
$ bash -c "command -v nvm"
$ bash -c "nvm --help"
bash: nvm: command not found
#!/bin/bash
a=${apt-cache show program}
if [[ $a == 0 ]]
then
echo "the program doesn't exist"
else
echo "the program exists"
fi
#program is not literal, you can change it to the program's name you want to check
The hash-variant has one pitfall: On the command line you can for example type in
one_folder/process
to have process executed. For this the parent folder of one_folder must be in $PATH. But when you try to hash this command, it will always succeed:
hash one_folder/process; echo $? # will always output '0'
I second the use of "command -v". E.g. like this:
md=$(command -v mkdirhier) ; alias md=${md:=mkdir} # bash
emacs="$(command -v emacs) -nw" || emacs=nano
alias e=$emacs
[[ -z $(command -v jed) ]] && alias jed=$emacs
I had to check if Git was installed as part of deploying our CI server. My final Bash script was as follows (Ubuntu server):
if ! builtin type -p git &>/dev/null; then
sudo apt-get -y install git-core
fi
To mimic Bash's type -P cmd, we can use the POSIX compliant env -i type cmd 1>/dev/null 2>&1.
man env
# "The option '-i' causes env to completely ignore the environment it inherits."
# In other words, there are no aliases or functions to be looked up by the type command.
ls() { echo 'Hello, world!'; }
ls
type ls
env -i type ls
cmd=ls
cmd=lsx
env -i type $cmd 1>/dev/null 2>&1 || { echo "$cmd not found"; exit 1; }
If there isn't any external type command available (as taken for granted here), we can use POSIX compliant env -i sh -c 'type cmd 1>/dev/null 2>&1':
# Portable version of Bash's type -P cmd (without output on stdout)
typep() {
command -p env -i PATH="$PATH" sh -c '
export LC_ALL=C LANG=C
cmd="$1"
cmd="`type "$cmd" 2>/dev/null || { echo "error: command $cmd not found; exiting ..." 1>&2; exit 1; }`"
[ $? != 0 ] && exit 1
case "$cmd" in
*\ /*) exit 0;;
*) printf "%s\n" "error: $cmd" 1>&2; exit 1;;
esac
' _ "$1" || exit 1
}
# Get your standard $PATH value
#PATH="$(command -p getconf PATH)"
typep ls
typep builtin
typep ls-temp
At least on Mac OS X v10.6.8 (Snow Leopard) using Bash 4.2.24(2) command -v ls does not match a moved /bin/ls-temp.
My setup for a Debian server:
I had the problem when multiple packages contained the same name.
For example apache2. So this was my solution:
function _apt_install() {
apt-get install -y $1 > /dev/null
}
function _apt_install_norecommends() {
apt-get install -y --no-install-recommends $1 > /dev/null
}
function _apt_available() {
if [ `apt-cache search $1 | grep -o "$1" | uniq | wc -l` = "1" ]; then
echo "Package is available : $1"
PACKAGE_INSTALL="1"
else
echo "Package $1 is NOT available for install"
echo "We can not continue without this package..."
echo "Exitting now.."
exit 0
fi
}
function _package_install {
_apt_available $1
if [ "${PACKAGE_INSTALL}" = "1" ]; then
if [ "$(dpkg-query -l $1 | tail -n1 | cut -c1-2)" = "ii" ]; then
echo "package is already_installed: $1"
else
echo "installing package : $1, please wait.."
_apt_install $1
sleep 0.5
fi
fi
}
function _package_install_no_recommends {
_apt_available $1
if [ "${PACKAGE_INSTALL}" = "1" ]; then
if [ "$(dpkg-query -l $1 | tail -n1 | cut -c1-2)" = "ii" ]; then
echo "package is already_installed: $1"
else
echo "installing package : $1, please wait.."
_apt_install_norecommends $1
sleep 0.5
fi
fi
}