是否可以捕获在 python 中的 try 块中引发的所有警告？

Question

I have some code that's schematically along the lines of:

from logging import warning

def complex_function():
    # Do some stuff
    raise Warning("blah") 
    # Do some more stuff
    raise Warning("Blah again") 

try: 
    complex_function()
except Warning as e: 
    warning(e) 

This results in:

WARNING:root:blah

I would like to catch all warnings raised, and log them. In my code, such warnings sometimes come from 3rd party libraries, so it is not practical to modify the warnings in place to use logging.warning , and I also want to store the warning information so that I can return some version of that information via an API.

Is there a way for me to do something like this that catches all warnings, and loops over them?

edit

Too late, I'm realising that I'm raising warnings wrong in the example above, and complex_function should be something long the lines of:

def complex_function():
    # Do some stuff
    warnings.warn("blah") 
    # Do some more stuff
    warnings.warn("Blah again", UnknownWarningType)

And I think I can catch these with warnings.catch_warnings

Answer 1

Were you expecting something like the following:

import warnings
warnings.filterwarnings('error')


def warning_func():
    print('hello')
    warnings.warn(Warning('Warn1'))
    print('hi')
    warnings.warn(Warning('Warn2'))


with warnings.catch_warnings(record=True) as w:
    warnings.simplefilter("always")
    warning_func()
    print(w)