Re: Re: Cronjob
| From: | Daren Cotter | Date: | Wed, 25 Sep 2002 18:32:34 +0000 |
| Subject: | Re: Re: Cronjob | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-117609@lists.php.net to get a copy of this message | ||
Jason,
That's exactly what I'm trying to do, and it's not
working:
My Script:
#!/usr/bin/php -f
<?php
$test = $argv[1];
print $test;
$demo = "This Works";
print $demo;
?>
Running:
./test.php blah
Yiels only "This Works", but not "blah".
I'm using version 4.0.6
--- Jason Young <jason1@erols.com> wrote:
> I can't say I'm really too familiar with the php
> commandline..
>
> You're using /usr/bin/php (or equivalent) and
> attempting to use your
> exec() that way?
>
> If you do 'php -?' you'll get a list of commands
> that you can use, and I
> don't see a way to pass cmdline arguments as
> variables..
>
> Having said that, I Just went and looked further
> into it.. if I make a
> test script, and at the top I put:
> $hi = $argv[1];
>
> then $hi becomes whatever you've specified as the
> first argument.. I'm
> assuming this is what you want?
>
> To clarify:
> phpfile.php contains:
> <?
> $hi = $argv[1];
> echo $hi;
> ?>
>
> Running the command "php -f phpfile.php test"
> returns "test"
>
> Does this help at all??
>
> -Jason
>
> Daren Cotter wrote:
> > Jason,
> >
> > I'm not using a web script any longer, I'm using
> > command-line (I determined that it is installed on
> the
> > server).
> >
> > I read about $argc and $argv, but when I call the
> > script passing two arguments, both $argc and $argv
> are
> > blank. Is this a php.ini setting I need to change
> or
> > somethign?
> >
> > --- Jason Young <jason1@erols.com> wrote:
> >
> >>Sorry to butt in :)
> >>
> >>Arguments to web scripts are done in the format:
> >>page.php?arg1=data1&arg2=data2
> >>
> >>So you would use that full string as the lynx
> path.
> >>
> >>Hope this helps :)
> >>-Jason
> >>
> >>Daren Cotter wrote:
> >>
> >>>Thanks for the info Chris, it works!
> >>>
> >>>How do I pass arguments to the script? I'm
> >>
> >>assuming
> >>
> >>>it'd just be:
> >>>
> >>>test.php arg1 arg2
> >>>
> >>>The stuff I've read says $argc should be the
> count
> >>
> >>of
> >>
> >>>the # of arguments, and $argv should be an array
> >>>holding them...but when I do a simple:
> >>>print "# of Arguments: $argc\n";
> >>>It prints nothing, not even 0
> >>>
> >>>
> >>>--- Chris Hewitt <g0pae@manordat.demon.co.uk>
> >>
> >>wrote:
> >>
> >>>>>On Wed, 25 Sep 2002, Daren Cotter wrote:
> >>>>>
> >>>>>
> >>>>>
> >>>>>>My problem, is that I absolutely NEED to run a
> >>>>>
> >>PHP
> >>
> >>>>>>script using crontab. The script needs to send
> >>>>>>numerous queries to a database every hour. Is
> >>>>>
> >>>>there
> >>>>
> >>>>
> >>>>>>any way I can accomplish this, directly or
> >>>>>
> >>>>indirectly?
> >>>>
> >>>>Are you sure its not already there? Commonly in
> >>>>/usr/bin. Try a "which
> >>>>php" and see if it finds anything?
> >>>>
> >>>>HTH
> >>>>Chris
> >>>>
> >>>
> >>>
> >>>
>
>>>__________________________________________________
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> >>
> >>
> >>--
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> >>
> >
> >
> >
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>
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