[英]Python Mocking - How to mock Google's storage.client?
Could anybody help with GCP API mocking?有人可以帮助 GCP API 模拟吗? Here's the function
func.py
:这是函数
func.py
:
import re
from google.cloud import storage
def is_all_log_entries_sink(sink):
storage_client = storage.Client()
if 'filter' not in sink and 'storage.googleapis.com' in sink.get('destination', ''):
bucket_name = re.search(r'^storage.googleapis.com/([^/]+)$', sink.get('destination')).group(1)
bucket = storage_client.lookup_bucket(bucket_name)
if bucket is not None:
return True
return False
Here's the test:这是测试:
import mock
from mock import patch, MagicMock
with mock.patch('oauth2client.client.GoogleCredentials.get_application_default') as mock_method:
import func
@patch('func.storage_client')
def test_is_all_log_entries_sink(mock_obj):
mock_obj.lookup_bucket = MagicMock(return_value='bucket-name')
sink = {'name': 'testname', 'destination': 'storage.googleapis.com/bucket-name'}
assert func.is_all_log_entries_sink(sink) == 1
assert mock_obj.lookup_bucket.called
sink = {'name': 'testname', 'destination': 'storage.googleapis.com/bucket-name', 'filter': 'testfilter'}
assert func.is_all_log_entries_sink(sink) == 0
sink = {'name': 'testname', 'destination': 'storage.googleapis.com/bucket-name'}
mock_obj.lookup_bucket = MagicMock(return_value=None)
assert func.is_all_log_entries_sink(sink) == 0
When PyTest ran, the following error was received:当 PyTest 运行时,收到以下错误:
E google.auth.exceptions.DefaultCredentialsError: Could not automatically determine credentials. Please set GOOGLE_APPLICATION_CREDENTIALS or explicitly create credentials and
re-run the application.
I tried to mock Google's authentication, but was unsuccessful.我试图模拟 Google 的身份验证,但没有成功。 Any assistance would be appreciated.
任何援助将不胜感激。
One possible solution: to mock GOOGLE_APPLICATION_CREDENTIALS as env variable in os module.一种可能的解决方案:将 GOOGLE_APPLICATION_CREDENTIALS 模拟为 os 模块中的 env 变量。 I consider GOOGLE_APPLICATION_CREDENTIALS as dict, so mocking dict may help you.
我认为 GOOGLE_APPLICATION_CREDENTIALS 是 dict,所以嘲笑 dict 可能会帮助你。
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