How do you pass a url as a REST GET parameter in FLASK

Question

So this is my code.

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. how-to-pass-urls-as-parameters-in-a-get-request-within-python-flask-restplus

Changing what I originally had.

 api.add_resource(LiveStats, '/liveStats/<string:url>') # Route_2

to this

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.

Example if I try this

localhost:60933/liveStats/http://address/Statistics?NoLogo=1%26KSLive=1

I get this

Trying to get http://address/Statistics

so it has taken off ?NoLogo=1%26KSLive=1

How do you prevent this?

score 0 · Answer 1 · answered Feb 14 '20 at 16:58

All characters after the ? are considered parameters. From the docs:

To access parameters submitted in the URL (?key=value) you can use the args attribute:

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.

For a full list of methods and attributes of the request object, head over to the Request documentation.

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.

If you don't want to access the args individually, you can access the full query string:

request.query_string

Putting this all together I think this will work for you:

class LiveStats(Resource):
  def get(self, url):
    return "Trying to get " + url + request.query_string

How do you pass a url as a REST GET parameter in FLASK

1 Answers1