How to parse URL from Python3 Requests Exception

Question

I am using Python3 requests module, doing a try/except and catching a timeout error specifically. The except statement returns the exception as a class object. How can I parse out particular elements of the error, for example, the URL? This is Python 3.7.8 on Ubuntu 18.04 if that matters.

try:
    result = requests.get(domain, timeout=5)
except requests.exceptions.Timeout as Err:
    print(type(Err))
    print(Err)

Returns --

<class 'requests.exceptions.ConnectTimeout'>
HTTPSConnectionPool(host='10.10.10.92', port=443): Max retries exceeded with url: /custom.php (Caused by ConnectTimeoutError(<urllib3.connection.HTTPSConnection object at 0x7c16f14f73d1>, 'Connection to 10.10.10.92 timed out. (connect timeout=5)'))

I'd like to be able to grab just the "Max retries exceeded with url: /custom.php" section.

Does this answer your question? Python: Error message parsing — mrCopiCat
– mrCopiCat, Commented Aug 22, 2022 at 20:20
Not really, though it did give me something to look at that I had missed. I guess I could turn the whole thing into a string then use substring operations to extract the part I wanted. I might give that a try. — BoredTech
– BoredTech, Commented Aug 22, 2022 at 20:30

p e p · Accepted Answer · 2022-08-22 21:47:34Z

1

Timeout inherits from RequestException, which has a request property. So you don't have to parse anything, the url is available via (in your example) Err.request.url. The request is of type requests.Request.

edited Aug 22, 2022 at 21:47

answered Aug 22, 2022 at 21:32

p e p

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