[英]How do you pass a url as a REST GET parameter in FLASK
So this is my code.所以这是我的代码。
from flask import Flask, request
from flask_restful import Resource, Api
app = Flask(__name__)
app.config['DEBUG'] = True
api = Api(app)
# Make the WSGI interface available at the top level so wfastcgi can get it.
wsgi_app = app.wsgi_app
class Default(Resource):
def get(self, name):
"""Renders a sample page."""
return "Hello " + name
class LiveStats(Resource):
def get(self, url):
return "Trying to get " + url
# data = request.get(url)
# return data
api.add_resource(Default, '/default/<string:name>') # Route_1
api.add_resource(LiveStats, '/liveStats/<path:url>') # Route_2
if __name__ == '__main__':
import os
HOST = os.environ.get('SERVER_HOST', 'localhost')
try:
PORT = int(os.environ.get('SERVER_PORT', '5555'))
except ValueError:
PORT = 5555
app.run(HOST, PORT)
Now firstly this post helped a lot.现在首先这篇文章有很大帮助。 how-to-pass-urls-as-parameters-in-a-get-request-within-python-flask-restplus
how-to-pass-urls-as-parameters-in-a-get-request-within-python-flask-restplus
Changing what I originally had.改变我原来拥有的。
api.add_resource(LiveStats, '/liveStats/<string:url>') # Route_2
to this对此
api.add_resource(LiveStats, '/liveStats/<path:url>') # Route_2
got rid of 404 errors that I had but now I am noticing that it's not passing all of the url.摆脱了我遇到的 404 错误,但现在我注意到它没有传递所有的 url。
Example if I try this例如,如果我尝试这个
localhost:60933/liveStats/http://address/Statistics?NoLogo=1%26KSLive=1
I get this我明白了
Trying to get http://address/Statistics
so it has taken off ?NoLogo=1%26KSLive=1
所以它已经起飞了
?NoLogo=1%26KSLive=1
How do you prevent this?你如何防止这种情况?
All characters after the ?
?
之后的所有字符are considered parameters.被视为参数。 From the docs:
从文档:
To access parameters submitted in the URL (?key=value) you can use the args attribute:
要访问在 URL (?key=value) 中提交的参数,您可以使用 args 属性:
searchword = request.args.get('key', '')
We recommend accessing URL parameters with get or by catching the KeyError because users might change the URL and presenting them a 400 bad request page in that case is not user friendly.
我们建议使用 get 或通过捕获 KeyError 来访问 URL 参数,因为用户可能会更改 URL 并向他们呈现 400 错误的请求页面,在这种情况下对用户不友好。
For a full list of methods and attributes of the request object, head over to the Request documentation.
有关请求对象的方法和属性的完整列表,请转到请求文档。
Maybe you could encode the query string in a way that you can retrieve it as a single parameter on the back end, but not sure it would be useful.也许您可以对查询字符串进行编码,以便在后端将其作为单个参数进行检索,但不确定它是否有用。
If you don't want to access the args individually, you can access the full query string:如果您不想单独访问 args,则可以访问完整的查询字符串:
request.query_string
Putting this all together I think this will work for you:把这一切放在一起,我认为这对你有用:
class LiveStats(Resource):
def get(self, url):
return "Trying to get " + url + request.query_string
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