Find out if a command exists on POSIX system

I want to be able to tell if a command exists on any POSIX system from a shell script.

On Linux, I can do the following:

if which <command>; then
   ...snip...
fi

However, Solaris and MacOS which do not give an exit failure code when the command does not exist, they just print an error message to STDOUT.

Also, I recently discovered that the which command itself is not POSIX (see http://www.opengroup.org/onlinepubs/009695399/utilities/)

Any ideas?

---

command -v is a POSIX specified command that does what which does.

It is defined to to return >0 when the command is not found or an error occurs.

---

POSIX does say, “If a command is not found, the exit status shall be 127.” So you could do

<command>
if [ "${?}" = 127 ]; then
   <handle not found>
fi

When writing shell scripts, it's often permissible to require a bash shell ( #!/bin/bash ), because without arrays it's pretty much impossible to handle arguments and/or filenames with spaces correctly. In that case, the bash builtin type -p is equivalent to which, and because it is builtin, it is portable.

---

You could read the stdout/stderr of "which" into a variable or an array (using backticks) rather than checking for an exit code.

If the system does not have a "which" or "where" command, you could also grab the contents of the $PATH variable, then loop over all the directories and search for the given executable. That's essentially what which does (although it might use some caching/optimization of $PATH results).

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