Making logging.getLogger() simpler

Kev Dwyer kevin.p.dwyer at gmail.com
Sun Sep 19 18:35:10 CEST 2010


On Sun, 19 Sep 2010 02:35:15 +1000, Lie Ryan wrote:

> I was expecting this to work:
> 
>   import logging
>   logger = logging.getLogger(__name__)
>   logger.warn('this is a warning')
> 
> instead it produced the error:
> 
>   No handlers could be found for logger "__main__"
> 
> 
> However, if instead I do:
> 
>   import logging
>   logging.warn('creating logger')
>   logger = logging.getLogger(__name__)
>   logger.warn('this is a warning')
> 
> then it does work.
> 
> Is there any reason why getLogger()-created logger shouldn't
> automatically create a default handler?

Hello Lie,

Calling logging.warn(), or logging.debug() etc. before handlers 
have been assigned to the root logger will result in 
logging.basicConfig() being called automatically.  By default a 
StreamHandler will be created on the root logger and your logger 
inherits the StreamHandler.  So you can avoid the "No handlers..."
warning by calling logging.basicConfig() before your program
does any logging.

I don't know why getLogger() doesn't so something similar when
it's called.  Perhaps so that the logger is explicitly 
initialised with basic, file or dictConfig?

Cheers,

Kev





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