Python – How to disable logging while running unit tests in Python Django

djangologgingpythonunit testing

I am using a simple unit test based test runner to test my Django application.

My application itself is configured to use a basic logger in settings.py using:

logging.basicConfig(level=logging.DEBUG)

And in my application code using:

logger = logging.getLogger(__name__)
logger.setLevel(getattr(settings, 'LOG_LEVEL', logging.DEBUG))

However, when running unittests, I'd like to disable logging so that it doesn't clutter my test result output. Is there a simple way to turn off logging in a global way, so that the application specific loggers aren't writing stuff out to the console when I run tests?

Best Answer

logging.disable(logging.CRITICAL)

will disable all logging calls with levels less severe than or equal to CRITICAL. Logging can be re-enabled with

logging.disable(logging.NOTSET)

Related Solutions

Python – Where do the Python unit tests go

For a file module.py, the unit test should normally be called test_module.py, following Pythonic naming conventions.

There are several commonly accepted places to put test_module.py:

In the same directory as module.py.
In ../tests/test_module.py (at the same level as the code directory).
In tests/test_module.py (one level under the code directory).

I prefer #1 for its simplicity of finding the tests and importing them. Whatever build system you're using can easily be configured to run files starting with test_. Actually, the default unittest pattern used for test discovery is test*.py.

Python – How to safely create a nested directory in Python

On Python ≥ 3.5, use pathlib.Path.mkdir:

from pathlib import Path
Path("/my/directory").mkdir(parents=True, exist_ok=True)

For older versions of Python, I see two answers with good qualities, each with a small flaw, so I will give my take on it:

Try os.path.exists, and consider os.makedirs for the creation.

import os
if not os.path.exists(directory):
    os.makedirs(directory)

As noted in comments and elsewhere, there's a race condition – if the directory is created between the os.path.exists and the os.makedirs calls, the os.makedirs will fail with an OSError. Unfortunately, blanket-catching OSError and continuing is not foolproof, as it will ignore a failure to create the directory due to other factors, such as insufficient permissions, full disk, etc.

One option would be to trap the OSError and examine the embedded error code (see Is there a cross-platform way of getting information from Python’s OSError):

import os, errno

try:
    os.makedirs(directory)
except OSError as e:
    if e.errno != errno.EEXIST:
        raise

Alternatively, there could be a second os.path.exists, but suppose another created the directory after the first check, then removed it before the second one – we could still be fooled.

Depending on the application, the danger of concurrent operations may be more or less than the danger posed by other factors such as file permissions. The developer would have to know more about the particular application being developed and its expected environment before choosing an implementation.

Modern versions of Python improve this code quite a bit, both by exposing FileExistsError (in 3.3+)...

try:
    os.makedirs("path/to/directory")
except FileExistsError:
    # directory already exists
    pass

...and by allowing a keyword argument to os.makedirs called exist_ok (in 3.2+).

os.makedirs("path/to/directory", exist_ok=True)  # succeeds even if directory exists.

Best Answer

Related Solutions

Python – Where do the Python unit tests go

Python – How to safely create a nested directory in Python

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