Google API HTTP 异常处理不起作用 (python)

Question

我正在尝试捕获 HTTP 错误。我使用了以下选项： 1) googleapiclient.errors.HttpError 2) google.api_core.exceptions.NotFound

这些选项都不适合我。任何想法为什么？谢谢！

from pprint import pprint

from googleapiclient import discovery, errors
from oauth2client.client import GoogleCredentials
#from google.api_core.exceptions import NotFound

credentials = GoogleCredentials.get_application_default()

service = discovery.build('compute', 'v1', credentials=credentials)

# Project ID for this request.
project = 'aweqwfnwrueifubwerfbiu'  # TODO: Update placeholder value.

try:
    request = service.firewalls().list(project=project)
except errors.HttpError:
    pprint('Error!')

while request is not None:
    response = request.execute()

    for firewall in response['items']:
        # TODO: Change code below to process each `firewall` resource:
        pprint(firewall)

    request = service.firewalls().list_next(previous_request=request, previous_response=response)

Traceback (most recent call last):
  File "test.py", line 20, in <module>
    response = request.execute()
  File "/lib/python3.6/site-packages/googleapiclient/_helpers.py", line 130, in positional_wrapper
    return wrapped(*args, **kwargs)
  File "/lib/python3.6/site-packages/googleapiclient/http.py", line 856, in execute
    raise HttpError(resp, content, uri=self.uri)
googleapiclient.errors.HttpError: <HttpError 404 when requesting https://compute.googleapis.com/compute/v1/projects/aweqwfnwrueifubwerfbiu/global/firewalls?alt=json returned "The resource 'projects/aweqwfnwrueifubwerfbiu' was not found">

Answer 1

试试这个来处理错误

from googleapiclient.errors import HttpError

但是，错误似乎并没有出现在您的异常捕获所在的位置。就在这里

while request is not None:
    response = request.execute()

这里没有

try:
    request = service.firewalls().list(project=project)
except errors.HttpError:
    pprint('Error!')

所以试着把它改成这个

while request is not None:
    try:
        response = request.execute()
    except errors.HttpError:
        pprint('Error!')