How to check if command exists in a shell script?
I am writing my first shell script. In my script I would like to check if a certain command exists, and if not, install the executable. How would I check if this command exists?
if #check that foobar command doesnt exist
then
#now install foobar
fi
In general, that depends on your shell, but if you use bash, zsh, ksh or sh (as provided by dash), the following should work:
if ! type "$foobar_command_name" > /dev/null; then
# install foobar here
fi
For a real installation script, you'd probably want to be sure that type doesn't return successfully in the case when there is an alias foobar . In bash you could do something like this:
if ! foobar_loc="$(type -p "$foobar_command_name")" || [[ -z $foobar_loc ]]; then
# install foobar here
fi
try using type :
type foobar
For example:
$ type ls
ls is aliased to `ls --color=auto'
$ type foobar
-bash: type: foobar: not found
This is preferable to which for a few reasons:
1) the default which implementations only support the -a option that shows all options, so you have to find an alternative version to support aliases
2) type will tell you exactly what you are looking at (be it a bash function or an alias or a proper binary).
3) type doesn't require a subprocess
4) type cannot be masked by a binary (for example, on a linux box, if you create a program called which which appears in path before the real which , things hit the fan. type , on the other hand, is a shell built-in [yes, a subordinate inadvertently did this once]
Check if a program exists from a Bash script covers this very well. In any shell script, you're best off running command -v $command_name for testing if $command_name can be run. In bash you can use hash $command_name , which also hashes the result of any path lookup, or type -P $binary_name if you only want to see binaries (not functions etc.)
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