Check if a program exists from a Bash script

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: http://pubs.opengroup.org/onlinepubs/9699919799/utilities/command.html

Note: type is POSIX compliant, but type -P is not.

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

It depends 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.

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 'git is not installed.' >&2
fi

Although if it's not executable, it may be better to special-case that, as that probably indicates a more serious issue.

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.

Try using:

test -x filename

or

[ -x filename ]

From the bash manpage under Conditional Expressions:

 -x file
          True if file exists and is executable.

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

I never did get the above solutions 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

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.

The which command might be useful. man which

It returns 0 if the executable is found, 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  exe-
          cutable

       2      if an invalid option is specified

Nice thing about which is that it figures out if the executable is available in the environment that which is run in - saves a few problems...

-Adam

For those interested, none of the methodologies above 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 distro):

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.

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'

Why not use Bash builtins if you can?

which programname

...

type -P programname

To mimic Bash's type -P cmd we can use 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; }

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

hash foo 2>/dev/null: works with zsh, bash, dash and ash.

type -p foo: it appears to work with zsh, bash and ash (busybox), but not dash (it interprets -p as an argument).

command -v foo: works with zsh, bash, dash, but not ash (busybox) (-ash: command: not found).

Also note that builtin is not available with ash and dash.

If there is no 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 10.6.8 using Bash 4.2.24(2) command -v ls does not match a moved /bin/ls-temp.

If you guys can't get the things above/below to work and 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

I use this because it's very easy:

if [ `LANG=C type example 2>/dev/null|wc -l` = 1 ];then echo exists;else echo "not exists";fi

or

if [ `LANG=C type example 2>/dev/null|wc -l` = 1 ];then
echo exists
else echo "not exists"
fi

It uses shell builtin and program echo status to stdout and nothing to stderr by the other hand if a command is not found, it echos status only to stderr.

I'd say there's no 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 is no superior to alias here.) There's no portable solution for it, and even a Bash-specific solution is rather tedious.

Note that solution like this will unconditionally reject alias ls='ls -F'

test() { command -v $1 | grep -qv alias }

Script copy paste to 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 printf "OK\n"; else printf "missing\n"; fi
done

Sample output:

latex     OK
pandoc    missing

Adjust the 10 to the maximum command length. Not automatic because I don't see a non verbose way to do it.

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

Hope this help someone else!

checkexists() {
    while [ -n "$1" ]; do
        [ -n "$(which "$1")" ] || echo "$1": command not found
        shift
    done
}

my setup for a debian server. i had a the problem when multiple packages contains 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
}

In case you want to check if a program exists and is really a program, not a bash built-in command, then command, type and hash are not appropriate for testing as they all return 0 exit status for built-in commands.

For example, there is the time program which offers more features than the time built-in command. To check if the program exists, I would suggest using which as in the following example:

# first check if the time program exists
timeProg=`which time`
if [ "$timeProg" = "" ]
then
  echo "The time program does not exist on this system."
  exit 1
fi

# invoke the time program
$timeProg --quiet -o result.txt -f "%S %U + p" du -sk ~
echo "Total CPU time: `dc -f result.txt` seconds"
rm result.txt

I couldn't get one of the solutions to work, but after editing it a little I came up with this. Which works for me:

dpkg --get-selections | grep -q linux-headers-$(uname -r)

if [ $? -eq 1 ]; then
        apt-get install linux-headers-$(uname -r)
fi