Inconsistent exit code from CLI? View as Markdown

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.

Using windows ...

@ECHO OFF
ECHO Calling PHP -v
PHP -v
ECHO %ERRORLEVEL%
ECHO Calling PHP -j
PHP -j
ECHO %ERRORLEVEL%

Both errorlevels are 0.

--
Richard Quadling
Twitter : EE : Zend
@RQuadling : e-e.com/M_248814.html : bit.ly/9O8vFY