我如何(安全地)将 Python 对象发送到我的 Flask API?
How do I (safely) send a Python object to my Flask API?
我目前正在尝试构建一个能够在 POST-请求中接收 python 对象的 Flask Web API。
我正在使用 Python 3.7.1 创建请求,Python 2.7 用于 运行 创建 API。 API 在我的本地机器上设置为 运行。我试图发送到我的 API 的对象是来自 sklearn.ensemble
的 RandomForestClassifier
对象,但这可能是多种对象类型中的任何一种。
到目前为止我已经尝试 json.dumps()
我的对象,但是这个对象不是 JSON 可序列化的。我也尝试过 pickle.dumps()
我的对象,但是当试图在 API 侧加载对象时这会导致错误。此外,由于 API 将接收来自匿名用户的请求,我担心对可能的恶意对象执行 pickle.loads()
。
这是有根据的担忧吗?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)
烧瓶API:
@app.route('/flask-api-function', methods=['POST'])
def flask_api_function():
model = pickle.loads(request.get_data())
在尝试使用 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
有什么好的又安全的方法吗?
当你在 python 3 中 pickle 对象时,传递协议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)
通常我会尝试找到一种方法将对象序列化为 JSON,因为 pickle 存在重大安全风险
我目前正在尝试构建一个能够在 POST-请求中接收 python 对象的 Flask Web API。
我正在使用 Python 3.7.1 创建请求,Python 2.7 用于 运行 创建 API。 API 在我的本地机器上设置为 运行。我试图发送到我的 API 的对象是来自 sklearn.ensemble
的 RandomForestClassifier
对象,但这可能是多种对象类型中的任何一种。
到目前为止我已经尝试 json.dumps()
我的对象,但是这个对象不是 JSON 可序列化的。我也尝试过 pickle.dumps()
我的对象,但是当试图在 API 侧加载对象时这会导致错误。此外,由于 API 将接收来自匿名用户的请求,我担心对可能的恶意对象执行 pickle.loads()
。
这是有根据的担忧吗?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)
烧瓶API:
@app.route('/flask-api-function', methods=['POST'])
def flask_api_function():
model = pickle.loads(request.get_data())
在尝试使用 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
有什么好的又安全的方法吗?
当你在 python 3 中 pickle 对象时,传递协议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)
通常我会尝试找到一种方法将对象序列化为 JSON,因为 pickle 存在重大安全风险