unexpected cli behavior

[yawnmoth]

i have the following two php scripts:

test.php:
<?
print system('./test-cli.php');
?>

test-cli.php:
#!/usr/local/bin/php -q
<?
print "hello, world!";
?>

when i run test.php from a webserver (test-cli.php is in the same
directory as test.php) i get the following output:

Usage: php [options] [-f] [args...] php [options] -r [args...] php
[options] [-- args...] -a Run interactively -c | Look for php.ini file
in this directory -n No php.ini file will be used -d foo[=bar] Define
INI entry foo with value 'bar' -e Generate extended information for
debugger/profiler -f Parse . -h This help -i PHP information -l Syntax
check only (lint) -m Show compiled in modules -r Run PHP without using
script tags -s Display colour syntax highlighted source. -v Version
number -w Display source with stripped comments and whitespace. -z
Load Zend extension . args... Arguments passed to script. Use -- args
when first argument starts with - or script is read from stdin starts
with - or script is read from stdin

so... what does this mean, and how do i fix it? any help would be
appreciated - thanks! :)

[Erich Musick]

Re: unexpected cli behavior

What if, in test.php you tried:

<?
print system('/usr/local/bin/php -q ./test-cli.php');
?>



yawnmoth wrote:[color=blue]
> i have the following two php scripts:
>
> test.php:
> <?
> print system('./test-cli.php');
> ?>
>
> test-cli.php:
> #!/usr/local/bin/php -q
> <?
> print "hello, world!";
> ?>
>
> when i run test.php from a webserver (test-cli.php is in the same
> directory as test.php) i get the following output:
>
> Usage: php [options] [-f] [args...] php [options] -r [args...] php
> [options] [-- args...] -a Run interactively -c | Look for php.ini file
> in this directory -n No php.ini file will be used -d foo[=bar] Define
> INI entry foo with value 'bar' -e Generate extended information for
> debugger/profiler -f Parse . -h This help -i PHP information -l Syntax
> check only (lint) -m Show compiled in modules -r Run PHP without using
> script tags -s Display colour syntax highlighted source. -v Version
> number -w Display source with stripped comments and whitespace. -z
> Load Zend extension . args... Arguments passed to script. Use -- args
> when first argument starts with - or script is read from stdin starts
> with - or script is read from stdin
>
> so... what does this mean, and how do i fix it? any help would be
> appreciated - thanks! :)[/color]

--
In Christ,

Erich Musick

http://erichmusick.com

In the same way, let your light shine before others, so that they may
see your good works and give glory to your Father who is in heaven. --
Matthew 5:16 NKJV

[yawnmoth]

Re: unexpected cli behavior

Erich Musick <erich.musick@s bcglobal.net> wrote in message news:<yEzmd.96$ rp2.68@newssvr3 1.news.prodigy. com>...[color=blue]
> What if, in test.php you tried:
>
> <?
> print system('/usr/local/bin/php -q ./test-cli.php');
> ?>[/color]

would that call the cli version, though? 'cause if it didn't,
wouldn't i be risking a so-called fork bomb, whereby a bunch of
processes are created, but none are closed?

[2metre]

Re: unexpected cli behavior

yawnmoth wrote:[color=blue]
> Erich Musick <erich.musick@s bcglobal.net> wrote in message news:<yEzmd.96$ rp2.68@newssvr3 1.news.prodigy. com>...
>[color=green]
>>What if, in test.php you tried:
>>
>><?
>> print system('/usr/local/bin/php -q ./test-cli.php');
>>?>[/color]
>
>
> would that call the cli version, though? 'cause if it didn't,
> wouldn't i be risking a so-called fork bomb, whereby a bunch of
> processes are created, but none are closed?[/color]
if the version of PHP at /usr/local/bin/php is the cli version, then yes
you are calling the cli version.

if the version of PHP at /usr/local/bin/php is NOT the cli version, then
no, you are NOT calling the cli version!

[yawnmoth]

Re: unexpected cli behavior

2metre <2metre@xxxhers ham.net> wrote in message news:<cnhnp6$lt 2$1@titan.btint ernet.com>...[color=blue]
> yawnmoth wrote:[color=green]
> > Erich Musick <erich.musick@s bcglobal.net> wrote in message news:<yEzmd.96$ rp2.68@newssvr3 1.news.prodigy. com>...
> >[color=darkred]
> >> <snip>[/color][/color]
> if the version of PHP at /usr/local/bin/php is the cli version, then yes
> you are calling the cli version.
>
> if the version of PHP at /usr/local/bin/php is NOT the cli version, then
> no, you are NOT calling the cli version![/color]

how would i know if it's the cli version or not? i'm running this on
a website that's hosted by some hosting company, so it's not as if i
installed php myself...