Hi, When php is invoked from the command line with -h we get exit code 0 as expected:
-bash-3.2# php -h ; echo $? Usage: php [options] [-f] <file> [--] [args...] php [options] -r <code> [--] [args...] php [options] [-B <begin_code>] -R <code> [-E <end_code>] [--] [args...] ... --re <name> Show information about extension <name>. --ri <name> Show configuration for extension <name>. 0 However if we use (what I understand to be an) invalid option we also get 0: php -Z ; echo $? Usage: php [options] [-f] <file> [--] [args...] .... 0 I've checked a couple of other tools (ls and grep) and they return exit code 2 when the options aren't recognised. Found on 5.2.14 and 5.3.3. Thanks, John. -- John Carter Development Manager johnc...@cisco.com