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

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.

Answer 2

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]

In addition, this will fail if the command you are looking for has been defined as an alias.

Answer 3

Try using:

test -x filename

or

[ -x filename ]

From the Bash manpage under Conditional Expressions:

 -x file
          True if file exists and is executable.

Answer 4

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

Answer 5

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.

Answer 6

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