我想要做的是从网站获取并且如果该请求返回401,则重做我的身份验证摆动(可能已过期)并再试一次。但我不想第三次尝试,因为这将是我的身份验证摆弄错误的凭据。有没有人有一个很好的方法做到这一点,并没有涉及到适当的丑陋代码,理想情况下在python请求库中,但我不介意改变。
答案 0 :(得分:3)
我认为它不会比这更难看:
import requests
from requests.auth import HTTPBasicAuth
response = requests.get('http://your_url')
if response.status_code == 401:
response = requests.get('http://your_url', auth=HTTPBasicAuth('user', 'pass'))
if response.status_code != 200:
# Definitely something's wrong
答案 1 :(得分:1)
你可以将它包装在一个函数中并使用装饰器来评估响应并在401上重试auth。然后你只需要装饰任何需要这个重新验证逻辑的函数....
更新:
根据要求,一个代码示例。我担心这个是一段旧代码,基于Python 2,但你会明白这一点。这个将按照settings.NUM_PLATFORM_RETRIES
中的定义多次重试http调用,并在auth失败时调用refresh_token
。你可以调整用例和结果。
然后,您可以围绕方法使用此装饰器:
@retry_on_read_error
def some_func():
do_something()
def retry_on_read_error(fn):
"""
Retry Feed reads on failures
If a token refresh is required it is performed before retry.
This decorator relies on the model to have a refresh_token method defined, othewise it will fail
"""
@wraps(fn)
def _wrapper(self, *args, **kwargs):
for i in range(settings.NUM_PLATFORM_RETRIES):
try:
res = fn(self, *args, **kwargs)
try:
_res = json.loads(res)
except ValueError:
# not a json response (could be local file read or non json data)
return res
if 'error' in _res and _res['error']['status'] in (401, 400):
raise AccessRefusedException(_res['error']['message'])
return res
except (urllib2.URLError, IOError, AccessRefusedException) as e:
if isinstance(e, AccessRefusedException):
self.refresh_token()
continue
raise ApiRequestFailed(
"Api failing, after %s retries: %s" % (settings.NUM_PLATFORM_RETRIES, e), args, kwargs
)
return _wrapper
答案 2 :(得分:0)
您可以使用类似的东西
library(broom)
contrived_data %>%
group_by(subgroup) %>%
summarize(avg = mean(value),
std_dev = sd(value),
out = list(tidy(ks.test(value, "pnorm", mean = 5, sd = 1)))) %>%
unnest_wider(c(out))