You can map the "debug" key on the fabric.api.output dict to True,
like in this example:
from fabric.api import local, output
output['debug'] = True
def go():
local("echo hi", capture=False)
And you can also use the --show=debug command-line argument.
I don't think you can control this from .fabricrc, but I could be wrong.
On Mon, Dec 7, 2009 at 10:04 PM, Marco Rogers <marco.rog...@gmail.com> wrote:
> I'm recently getting familiar with context managers in python. I'd like to
> be able to set the debug output option globally for my fabfile script
> instead of doing "with show('debug'):" in every task. Ideally I could set
> it in the .fabricrc, but if I can update the global context at the top of
> the script that'd be just as well. Are either of these currently possible
> and if so, what's the syntax?
> Sorry if I missed this in the docs, but I went over them pretty thoroughly
> and it doesn't seem to address this particular task.
> Thanks, really liking fabric
> :Marco
>
> --
> Marco Rogers
> marco.rog...@gmail.com
>
> Life is ten percent what happens to you and ninety percent how you respond
> to it.
> - Lou Holtz
>
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> Fab-user@nongnu.org
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>
>
--
Venlig hilsen / Kind regards,
Christian Vest Hansen.
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