I'm having some difficulty in getting a debug session started from the command line, it works fine through a browser and if I put in a DebugBreak() in the script, even the command line will break into a debugger.

I'm running it as such:
php test.php DBGSESSID=1@127.0.0.1:7869;d=1

php -v
PHP 5.2.2 (cli) (built: May 2 2007 19:18:26)
Copyright (c) 1997-2007 The PHP Group
Zend Engine v2.2.0, Copyright (c) 1998-2007 Zend Technologies
with DBG v3.1.11, (C) 2000,2007, by Dmitri Dmitrienko
with Zend Extension Manager v1.2.0, Copyright (c) 2003-2007, by Zend Technologies
with Zend Optimizer v3.2.8, Copyright (c) 1998-2007, by Zend Technologies


Any ideas what may be up?

I'm not sure I got you right.
Please provide more details on what you're trying to achive.

I'm running into this problem. I'm trying to start a command line script from a linux command prompt with the following:

php test.php Profile=dev DBGSESSID=1@72.84.10.104:7869;d=1,p=1

But its not connecting to my windows box, I'm watching the PHP DBG Listener running on my local machine.

I can put in a DebugBreak('1@72.84.10.104:7869;d=1,p=1'); and it starts the debug session fine.

Any idea why I can't start this from the command line? I read on some post this was allowed as a command line argument.

Thanks!

is it CLI or CGI version of php you're trying to debug?
please check with php -i output

It's CLI:

PHP 5.2.5 (cli) (built: Dec 11 2007 10:20:33)
Copyright (c) 1997-2007 The PHP Group
Zend Engine v2.2.0, Copyright (c) 1998-2007 Zend Technologies
with DBG v3.2.10, (C) 2000,2007, by Dmitri Dmitrienko
with Zend Extension Manager v1.2.2, Copyright (c) 2003-2007, by Zend Technologies

I just checked and it worked fine for me.
Please check if $_SERVER['argv'] array contains DBGSESSID variable with appropriate content you specified in the command line.