[Solved] Passing variable to php cron job.

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WorldCom
Forum Commoner
Posts: 45
Joined: Sat Jun 24, 2006 8:14 am
Location: Ontario, Canada

[Solved] Passing variable to php cron job.

Post by 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:

Code: Select all

php -q /home/mysite/public_html/cron/update_cron_22.php id=22
Any help appreciated.
Last edited by WorldCom on Sun Aug 12, 2007 3:02 pm, edited 1 time in total.
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VladSun
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Posts: 4313
Joined: Wed Jun 27, 2007 9:44 am
Location: Sofia, Bulgaria

Post by 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.
There are 10 types of people in this world, those who understand binary and those who don't
WorldCom
Forum Commoner
Posts: 45
Joined: Sat Jun 24, 2006 8:14 am
Location: Ontario, Canada

Post by WorldCom »

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