Re: Re: Cronjob

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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