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Date: Tue, 26 Oct 2010 11:59:59 +0100
Message-ID: <AANLkTimJEq6e62AtwE74Q=JsY7f1YJ9zs-Cj+vZfGrdj@mail.gmail.com>
To: John Carter <johncart@cisco.com>
Cc: internals@lists.php.net
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Subject: Re: [PHP-DEV] Inconsistent exit code from CLI?
From: rquadling@gmail.com (Richard Quadling)

On 26 October 2010 09:29, John Carter <johncart@cisco.com> wrote:
> 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...]
> =C2=A0 =C2=A0 =C2=A0 php [options] -r <code> [--] [args...]
> =C2=A0 =C2=A0 =C2=A0 php [options] [-B <begin_code>] -R <code> [-E <end_c=
ode>] [--]
> [args...]
> ...
> =C2=A0--re <name> =C2=A0 =C2=A0 =C2=A0Show information about extension <n=
ame>.
> =C2=A0--ri <name> =C2=A0 =C2=A0 =C2=A0Show configuration for extension <n=
ame>.
>
> 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.

--=20
Richard Quadling
Twitter : EE : Zend
@RQuadling : e-e.com/M_248814.html : bit.ly/9O8vFY