include path and cron

Question

I'm running a cronjob that calls a php script. I get "failed to open stream" when the file is invoked by cron. When I cd to the directory and run the file from that location, all is well. Basically, the include_once() file that I want to include is two directories up from where the php script resides.

Can someone please tell me how I can get this to work from a cronjob?

Answer 1

cron is notorious for starting with a minimal environment. Either:

• have your script set up it's own environment;
• have a special cron script which sets up the environment then calls your script; or
• set up the environment within crontab itself.

An example of the last (which is what I tend to use if there's not too many things that need setting up) is:

0 5 * * * (export PATH = /mydir:$PATH ; myexecutable )

Answer 2

you need to see what is the path that the cron run from.

 echo pathinfo($_SERVER["PATH_TRANSLATED"]);

according to this do the include

include $path_parts['dirname']."/myfile.php";

Answer 3

There are multiple ways to do this: You could cd into the directory in your cron script:

cd /path/to/your/dir && php file.php

Or point to the correct include file relative to the current script in PHP:

include dirname(__FILE__) . '/../../' . 'includedfile.php';