How to correctly handle no network access when using Python requests.get()

Question

I'm retrieving data from a URL using reqests.get(). My device is connected by wifi. Occasionally, there is no internet access and my appliaction fails with the following error:

OSError: [Errno 101] Network is unreachable
During handling of the above exception, another exception occurred:
....
During handling of the above exception, another exception occurred:
...
During handling of the above exception, another exception occurred:

My requests.get() code is surrounded by the a try block:

 try:
 response = requests.get(url, timeout=10)
 except requests.Timeout as e:
 print(e)
 return 0
 if response.status_code == 200:
 my_data = response.json()
 # do some stuff
 return my_data
 else:
 return 0

How do I alter the try block so that it returns 0 if the response status code is not 200, and also returns 0 if there is no network access?

• wim: Why would you want to return 0 in this case? Just let the exception unhandled. Python is not C ...
• jackal: Just add another except clause

Answer 1

All exceptions explicitly raised by requests inherit from requests.exceptions.RequestException. Therefore, you can simply use response.raise_for_status() and handle all exceptions in a try-except block.

import requests 
def _get(url):
 try: 
 response = requests.get(url, timeout=10)
 response.raise_for_status()
 return response.json()
 except requests.exceptions.RequestException as exc:
 return 0 

• wim: raise_for_status method itself just checks the status_code and raises an exception if the code is >= 400. Use it when you want the exception to bubble up. If you're just going to handle that exception and return 0, that's a pointless indirection.