Python urllib2: Reading content body even during HTTPError exception?

Question

I’m using urllib2 to fetch a a page via HTTP. Sometimes the resource throws a HTTP error 400 (Bad Request) when my request contains an error. However, that response also contains an XML element that gives a detailed error message. It would be very handy to be able to see that error rather than just the HTTPError exception returned by urllib2.

How do I return the document contents in spite of the exception?

Asked By: jamieb

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Source

Answer 1

You can treat the error as a response.

http://www.voidspace.org.uk/python/articles/urllib2.shtml#httperror

When an error is raised the server
responds by returning an HTTP error
code and an error page. You can use
the HTTPError instance as a response
on the page returned. This means that
as well as the code attribute, it also
has read, geturl, and info, methods.

Answered By: Gareth Simpson

Answer 2

import urllib2
try:
    request = urllib2.Request('http://www.somesite.com')
    response = urllib2.urlopen(req)
except urllib2.HTTPError as e:
    error_message = e.read()
    print error_message

Answered By: Tendayi Mawushe

Answer 3

You can read the response message from the HTTPError exception.

Python3 example

import urllib.request

try:
    request = urllib.request.Request('http://httpstat.us/418', headers={'Accept': 'text/plain', 'User-Agent': ''})
    with urllib.request.urlopen(request) as page:
        print('success: ' + page.read().decode())
except urllib.error.HTTPError as httpError:
        error = httpError.read().decode()
        print('error: ' + error)

Answered By: Synck

Python urllib2: Reading content body even during HTTPError exception?

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