Python logging typeerror
14,500
Solution 1
logging.INFO
denotes an integer constant with value of 20
INFO Confirmation that things are working as expected.
What you need is logging.info
logging.info("test")
Solution 2
You are trying to call logging.INFO
, which is an integer constant denoting one of the pre-defined logging levels:
>>> import logging
>>> logging.INFO
20
>>> type(logging.INFO)
<type 'int'>
You probably wanted to use the logging.info()
function (note, all lowercase) instead:
Logs a message with level
INFO
on this logger. The arguments are interpreted as fordebug()
.
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Author by
FrUh
Updated on August 11, 2020Comments
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FrUh almost 4 years
Could you please help me, whats wrong.
import logging if (__name__ == "__main__"): logging.basicConfig(format='[%(asctime)s] %(levelname)s::%(module)s::%(funcName)s() %(message)s', level=logging.DEBUG) logging.INFO("test")
And I can't run it, I've got an error:
Traceback (most recent call last): File "/home/htfuws/Programming/Python/just-kidding/main.py", line 5, in logging.INFO("test") TypeError: 'int' object is not callable
Thank you very much.
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Matt almost 11 yearsIs this the whole code?
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FrUh almost 11 yearsthank you very much, I was using it in my previuos project and I was wondering why it doesn't work. AND I DID NOT NOTICE the LOWER CASE. Ah.
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FrUh almost 11 yearsthank you very much, I was using it in my previuos project and I was wondering why it doesn't work. AND I DID NOT NOTICE the LOWER CASE. Ah.
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Martijn Pieters almost 11 yearsAnd you did not notice the CAPS LOCK either, by the looks of it. :-P (And sorry, you can only mark one answer as accepted, thanks for the brief acceptance though!)
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alphazwest over 4 yearsmy god man, this has been the source of so much frustration for me.