[Solved] Passing variable to php cron job.

[WorldCom]

I'm trying to do something that should be so simple.

I want to pass a variable to a php cron job script.
This is the php script:

Code: Select all

#!/usr/bin/php
<?php
require('connect.php');
$id = (int)$_GET['id'];
$game_data = mysql_fetch_array(mysql_query("SELECT game_id FROM game_data WHERE week=id"));
$update = "UPDATE results_data SET status='Closed' WHERE results_id=$game_data[0]";
$update_info = mysql_query($update) or die ('Database Error: ' . mysql_error());
echo "Cron Job ran Successfully";
?>

This works perfectly if put the 'id' in directly and not pass it. However, trying to pass the data gives the 'Input File not found error'. Now, doing some searching I came across a thread that said you can't pass variables in the tradtional sense. eg $_GET So the syntax was to just use a space.

eg. script.php variable1=value

This worked in the fact that I now get a MySQL error because of not reading the 'id' variable.

So how do you read php variables with a space instead of a question mark into the script?

Heres the cron job command line:

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php -q /home/mysite/public_html/cron/update_cron_22.php id=22

Any help appreciated.

[VladSun]

http://www.phpbuilder.com/columns/darre ... hp3?page=2

You don't have to do this like GET (i.e. var=value) - instead:

Code: Select all

php -q /home/mysite/public_html/cron/update_cron_22.php 22

And you'll have in $argv[1] value of 22.

[WorldCom]

That's exactly what I was looking for.
Works great thanks