Clean way to handle file not found exception

Question

BACKGROUND

I have a function where i cache file from aws s3. I provide it the path on s3 and also the local path of where i want the cached file to live. Sometimes if the file does not exist on s3 it will return a 404 error. That makes sense but i want to handle it gracefully. What i want to do is if the file i am trying to cache does not exist I want my get_overview_stream function to return None.

CODE

def get_overview_stream(*, environment, proxy_key):
 s3_overview_file_path = f"s3://{FOO_TRACK_BUCKET}/{environment}"
 overview_file = f"{proxy_key}.csv"
 local_path = cache_directory(environment)
 try:
 cache.cache_file(s3_overview_file_path, local_path, overview_file)
 overview_file_cache = local_path / f"{proxy_key}.csv"
 return overview_file_cache.open("r")
 except botocore.exceptions.ClientError as e:
 if e.response["Error"]["Code"] == "404":
 return None
 else:
 raise

Attempted Solution Issue

I am very new to python so i am not sure if this the best way to handle this exception and if there is a more cleaner way. If so i would love to hear feedback.

• Carcigenicate: Honestly, I've never used to libraries involved here, but I don't see anything wrong with this approach. The exception handling seems fine to me.
• Maarten Fabré: The only thing I would change here is adding a docstring and type information. You could move the last 2 lines in the try block outside. raising a custom exception instead of returning the sentinel value None might also be an improvement, but that last point id a matter of taste /habit
• Maarten Fabré: Also, always open a file with a context manager instead of a bare open()
• Dinero: @MaartenFabré can you show example of what you mean by custom exception?

Answer 1

I would suggest not opening the file in your function but instead returning the path or None, such that the caller can open the file using a with statement.