How can I check if a program exists from a Bash script?

Question

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\

Answer 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.

Answer 2

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.

Answer 3

This will tell according to the location if the program exist or not:

    if [ -x /usr/bin/yum ]; then
        echo "This is Centos"
    fi

Answer 4

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

Answer 5

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
}

Answer 6

Late answer but this is what I ended up doing.

I just check if the command I execute returns an error code. If it returns 0 it means program is installed. Moreover you can use this to check the output of a script. Take for instance this script.

foo.sh

#!/bin/bash
echo "hello world"
exit 1 # throw some error code

Examples:

# outputs something bad... and exits
bash foo.sh $? -eq 0 || echo "something bad happened. not installed" ; exit 1

# does NOT outputs nothing nor exits because dotnet is installed on my machine
dotnet --version $? -eq 0 || echo "something bad happened. not installed" ; exit 1

Basically all this is doing is checking the exit code of a command run. the most accepted answer on this question will return true even if the command exit code is not 0.