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我需要消耗庫中定義的一些方法。 在我的模塊,我有如何訪問方法庫?
/mymodule/lib/mod.py
/mymodule/views.py
而我views.py
在我mod.py
我宣佈:
class ModClient(object):
"""REST client for Mod API"""
def __init__(self, client_id, secret, environment):
self.client_id = client_id
self.secret = secret
self.environment = environment
def _base_url(self):
base_url = ''
if self.environment == 'sandbox':
base_url = 'https://sandbox.mod.com'
elif self.environment == 'development':
base_url = 'https://development.mod.com'
elif self.environment == 'production':
base_url = 'https://production.mod.com'
return base_url
def _base_params(self):
params = {
'client_id': self.client_id,
'secret': self.secret
}
return params
def _parse_response(self, response):
result = response.json()
if response.status_code != 200:
raise ModClientException(message='HTTP status {}: {}'.format(response.status_code, result),
http_status=response.status_code,
error_type=result.get('error_type', None),
error_code=result.get('error_code', None))
return result
def get_accounts(self, access_token):
url = '{}/accounts/get'.format(self._base_url())
params = self._base_params()
params['access_token'] = access_token
response = requests.post(url, json=params)
return self._parse_response(response)
如何我可以從我view.py
訪問我的方法get_accounts
假設兩者都在同一個模塊中?
你可以在view.py中實例化'ModClient',然後調用'instance.get_accounts()'? –
JacobIRR
'mod.py','view.py'聽起來很像兩個不同的模塊給我!你的意思是它們在同一個包裏('mymodule.lib可能')? – schwobaseggl
yes,sorry,differente packages,'views.py'在'/ mymodule/views.py'中 – MrMins