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?
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 $PATH results).
POSIX does say, “If a command is not found, the exit status shall be 127.” So you could do
<command>
if [ "${?}" = 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.
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.