python - 如何(安全)将 Python 对象发送到我的 Flask API？

Question

我目前正在尝试构建一个 Flask Web API，它能够在 POST 请求中接收 python 对象。

我使用 Python 3.7.1 创建请求，使用 Python 2.7 运行 API。该 API 设置为在我的本地计算机上运行。我尝试发送到 API 的对象是来自 sklearn.ensemble 的 RandomForestClassifier 对象，但这可以是多种对象类型中的任何一种。

到目前为止，我已尝试 json.dumps() 我的对象，但该对象不可 JSON 序列化。我还尝试了 pickle.dumps() 我的对象，但这在尝试在 API 端加载对象时导致了错误。此外，由于 API 将接收来自匿名用户的请求，因此我担心对可能的恶意对象执行 pickle.loads()。

这是一种有根据的担忧吗？如果是这样，在 POST 请求中发送任何 python 对象的最佳方式是什么？

执行 POST 请求的脚本:

import requests
import pickle

url = "http://localhost:5000/flask-api-function"

# the object I want to send is the 'model' object
data = pickle.dumps(model)

r = requests.post(url,data=data)

Flask API:

@app.route('/flask-api-function', methods=['POST'])
def flask_api_function():

  model = pickle.loads(request.get_data())

当尝试使用 pickle 解码数据时，此设置实际上会导致错误:

Traceback (most recent call last):
  File "/Library/Python/2.7/site-packages/flask/app.py", line 2309, in __call__
return self.wsgi_app(environ, start_response)
  File "/Library/Python/2.7/site-packages/flask/app.py", line 2295, in wsgi_app
response = self.handle_exception(e)
  File "/Library/Python/2.7/site-packages/flask/app.py", line 1741, in handle_exception
reraise(exc_type, exc_value, tb)
  File "/Library/Python/2.7/site-packages/flask/app.py", line 2292, in wsgi_app
response = self.full_dispatch_request()
  File "/Library/Python/2.7/site-packages/flask/app.py", line 1815, in full_dispatch_request
rv = self.handle_user_exception(e)
  File "/Library/Python/2.7/site-packages/flask/app.py", line 1718, in handle_user_exception
reraise(exc_type, exc_value, tb)
  File "/Library/Python/2.7/site-packages/flask/app.py", line 1813, in full_dispatch_request
rv = self.dispatch_request()
  File "/Library/Python/2.7/site-packages/flask/app.py", line 1799, in dispatch_request
return self.view_functions[rule.endpoint](**req.view_args)
  File "/Users/larssuanet/Documents/enjins/dscs/flask_api.py", line 39, in store_model
model = pickle.loads(request.get_data())
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 1382, in loads
return Unpickler(file).load()
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 858, in load
dispatch[key](self)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 886, in load_proto
raise ValueError, "unsupported pickle protocol: %d" % proto
ValueError: unsupported pickle protocol: 3

有没有一种既好又安全的方法呢？

Answer 1

当您在 python 3 中 pickle 对象时，传递协议(protocol)关键字并将其设置为 2。这将确保它可以在 python 2 上工作。

import requests
import pickle

url = "http://localhost:5000/flask-api-function"

# the object I want to send is the 'model' object
data = pickle.dumps(model,protocol=2)

r = requests.post(url,data=data)

一般来说，我会尝试找到一种将对象序列化为 JSON 的方法，因为 pickle 存在重大安全风险