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Using https as standard with django project

I am learning django and trying to complete my first webapp.
I am using shopify api & boilder plate (starter code) and am having an issue with the final step of auth.
Specifically, the redirect URL -- it's using HTTP:// when it should NOT and I don't know how to change it..
#in my view
def authenticate(request):
shop = request.GET.get('shop')
print('shop:', shop)
if shop:
scope = settings.SHOPIFY_API_SCOPE
redirect_uri = request.build_absolute_uri(reverse('shopify_app_finalize')) #try this with new store url?
print('redirect url', redirect_uri) # this equals http://myherokuapp.com/login/finalize/
permission_url = shopify.Session(shop.strip()).create_permission_url(scope, redirect_uri)
return redirect(permission_url)
return redirect(_return_address(request))
Which is a problem because my app uses the Embedded Shopify SDK which causes this error to occur at the point of this request
Refused to frame 'http://my.herokuapp.com/' because it violates the following Content Security Policy directive: "child-src 'self' https://* shopify-pos://*". Note that 'frame-src' was not explicitly set, so 'child-src' is used as a fallback.
How do i change the URL to use HTTPS?
Thank you so much in advance. Please let me know if I can share any other details but my code is practically identical to that starter code

This is what the Django doc says about build_absolute_uri:
Mixing HTTP and HTTPS on the same site is discouraged, therefore
build_absolute_uri() will always generate an absolute URI with the
same scheme the current request has. If you need to redirect users to
HTTPS, it’s best to let your Web server redirect all HTTP traffic to
HTTPS.
So you can do two things:
Make sure your site runs entirely on HTTPS (preferred option): Setup your web server to use HTTPS, see the Heroku documentation on how to do this. Django will automatically use HTTPS for request.build_absolute_uri if the incoming request is on HTTPS.
I'm not sure what gets passed in the shop parameter but if it contains personal data I'd suggest to use HTTPS anyway.
Create the URL yourself:
url = "https://{host}{path}".format(
host = request.get_host(),
path = reverse('shopify_app_finalize'))
But you will still need to configure your server to accept incoming HTTPS requests.

How to pass URLs as parameters in a GET request within python flask (restplus)?

I am creating a REST API using python flask. The API is ready and works on port number 8000 of my localhost. Now I intend to give this REST API a user friendly interface for which I decided to go with python - restplus. I thought of calling this service (running on 8000) internally from swagger application running on 5000
I was able to create the basic structure of the API (Swagger). The code for which looks like this:
import flask
from flask import Flask, request
from flask_restplus import Resource, Api
app = Flask(__name__)
api = Api(app)
#api.route('/HybridComparator/<string:url2>/<string:url1>')
class HybridComparator(Resource):
def get(self, url1, url2):
print(url1)
print(url2)
return url1 + ' ' + url2
if __name__ == '__main__':
app.run(debug=True)
The application as a whole runs seamlessly (with random strings as parameters) on port 5000. But when the URLs I pass are actual links, the application returns a response of 404 - Not found. Further to my investigation I realized the culprit being '/' embedded within the links I try to provide. Is there a way to handle URLs in particular?
Should I encode them before sending a request. (This will make my parameters look ugly). Is there something I am missing?

This is an entirely old question and I am sure you solved your problem by now.
But for new searchers, this may come in handy;
replace <string:url2>/<string:url1> with <path:url2>/<path:url1>

it seems that :
#api.route('/HybridComparator/<path:url2>/<path:url1>')
should fix it ,it fixes the 404 but i am getting only "http:/" part of the param

Accesing ASANA data using python requests

First of all, I'm not a Python guru as you can probably tell... So here we go.
I'm trying to use Asana's API to pull data with Python requests (Projects, tasks, etc) and doing the authentication using Oauth 2.0... I've been trying to find a simple python script to have something to begin with but I haven't had any luck and I can't find a decent and simple example!
I already created the app and got my client_secret and client_secret. But I don't really know where or how to start... Could anybody help me please?
import sys, os, requests
sys.path.append(os.path.dirname(os.path.dirname(__file__)))
import asana
import json
from six import print_
import requests_oauthlib
from requests_oauthlib import OAuth2Session
client_id=os.environ['ASANA_CLIENT_ID'],
client_secret=os.environ['ASANA_CLIENT_SECRET'],
# this special redirect URI will prompt the user to copy/paste the code.
# useful for command line scripts and other non-web apps
redirect_uri='urn:ietf:wg:oauth:2.0:oob'
if 'ASANA_CLIENT_ID' in os.environ:
#Creates a client with previously obtained Oauth credentials#
client = asana.Client.oauth(
#Asana Client ID and Secret, set as a Windows environments to avoid hardcoding variables into the script#
client_id=os.environ['ASANA_CLIENT_ID'],
client_secret=os.environ['ASANA_CLIENT_SECRET'],
# this special redirect URI will prompt the user to copy/paste the code.
# useful for command line scripts and other non-web apps
redirect_uri='urn:ietf:wg:oauth:2.0:oob'
)
print ("authorized=", client.session.authorized)
# get an authorization URL:
(url, state) = client.session.authorization_url()
try:
# in a web app you'd redirect the user to this URL when they take action to
# login with Asana or connect their account to Asana
import webbrowser
webbrowser.open(url)
except Exception as e:
print_("Open the following URL in a browser to authorize:")
print_(url)
print_("Copy and paste the returned code from the browser and press enter:")
code = sys.stdin.readline().strip()
# exchange the code for a bearer token
token = client.session.fetch_token(code=code)
#print_("token=", json.dumps(token))
print_("authorized=", client.session.authorized)
me = client.users.me()
print "Hello " + me['name'] + "\n"
params = {'client_id' : client_id, 'redirect_uri' : redirect_uri, 'response_type' : token,}
print_("*************** Request begings *******************"+"\n")
print_("r = requests.get('https://app.asana.com/api/1.0/users/me)" + "\n")
r = requests.get('https://app.asana.com/api/1.0/users/me', params)
print_(r)
print_(r.json)
print_(r.encoding)
workspace_id = me['workspaces'][0]['id']
print_("My workspace ID is" + "\n")
print_(workspace_id)
print_(client.options)
I'm not sure how to use the requests lib with Asana. Their python doc did not help me. I'm trying to pull the available projects and their code colours so I can later plot them into a web browser (For a high-level view of the different projects and their respective colours - Green, yellow or red)
When I introduce the url (https://app.asana.com/api/1.0/users/me) into a browser, it gives me back a json response with the data, but when I try to do the same with the script, it gives me back a 401 (not authorized) response.
Does anybody know what I'm missing / doing wrong?
Thank you!!!

I believe the issue is that the Requests library is a lower level library. You would need to pass all of the parameters to your requests.
Is there a reason you are not exclusively using the Asana Python client library to make requests? All of the data you are looking to fetch from Asana (projects, tasks, etc.) are accessible using the Asana Python library. You will want to look in the library to find the methods you need. For example, the methods for the tasks resource can be found here. I think this approach will be easier (and less error-prone) than switching between the Asana lib and the Requests lib. The Asana lib is actually built on top of Requests (as seen here).

Change URL to another URL using mitmproxy

I am trying to redirect one page to another by using mitmproxy and Python. I can run my inline script together with mitmproxy without issues, but I am stuck when it comes to changing the URL to another URL. Like if I went to google.com it would redirect to stackoverflow.com
def response(context, flow):
print("DEBUG")
if flow.request.url.startswith("http://google.com/"):
print("It does contain it")
flow.request.url = "http://stackoverflow/"
This should in theory work. I see http://google.com/ in the GUI of mitmproxy (as GET) but the print("It does contain it") never gets fired.
When I try to just put flow.request.url = "http://stackoverflow.com" right under the print("DEBUG") it won't work neither.
What am I doing wrong? I have also tried if "google.com" in flow.request.url to check if the URL contains google.com but that won't work either.
Thanks

The following mitmproxy script will
Redirect requesto from mydomain.com to newsite.mydomain.com
Change the request method path (supposed to be something like /getjson? to a new one `/getxml
Change the destination host scheme
Change the destination server port
Overwrite the request header Host to pretend to be the origi
import mitmproxy
from mitmproxy.models import HTTPResponse
from netlib.http import Headers
def request(flow):
if flow.request.pretty_host.endswith("mydomain.com"):
mitmproxy.ctx.log( flow.request.path )
method = flow.request.path.split('/')[3].split('?')[0]
flow.request.host = "newsite.mydomain.com"
flow.request.port = 8181
flow.request.scheme = 'http'
if method == 'getjson':
flow.request.path=flow.request.path.replace(method,"getxml")
flow.request.headers["Host"] = "newsite.mydomain.com"

You can set .url attribute, which will update the underlying attributes. Looking at your code, your problem is that you change the URL in the response hook, after the request has been done. You need to change the URL in the request hook, so that the change is applied before requesting resources from the upstream server.

Setting the url attribute will not help you, as it is merely constructed from underlying data. [EDIT: I was wrong, see Maximilian’s answer. The rest of my answer should still work, though.]
Depending on what exactly you want to accomplish, there are two options.
(1) You can send an actual HTTP redirection response to the client. Assuming that the client understands HTTP redirections, it will submit a new request to the URL you give it.
from mitmproxy.models import HTTPResponse
from netlib.http import Headers
def request(context, flow):
if flow.request.host == 'google.com':
flow.reply(HTTPResponse('HTTP/1.1', 302, 'Found',
Headers(Location='http://stackoverflow.com/',
Content_Length='0'),
b''))
(2) You can silently route the same request to a different host. The client will not see this, it will assume that it’s still talking to google.com.
def request(context, flow):
if flow.request.url == 'http://google.com/accounts/':
flow.request.host = 'stackoverflow.com'
flow.request.path = '/users/'
These snippets were adapted from an example found in mitmproxy’s own GitHub repo. There are many more examples there.
For some reason, I can’t seem to make these snippets work for Firefox when used with TLS (https://), but maybe you don’t need that.

In Flask is there a way to ignore a request for a route that doesn't exist

I'm writing an app using Flask.
I have a set of routes and they work.
What I want to do on the client side is to ignore any requests to invalid URLs. That is I do not want to render any 404/error pages in the app. I would like an alert that says the URL is invalid and for the browser to simply stay on the same page.
I don't want to be checking the URLs in JavaScript on the client, as this would expose them.
I have a route which responds correctly to unknown URLs:
#app.errorhandler(404)
def non_existant_route(error):
return jsonify({"no":"such page"})
If I delete the return statement I get a 500 error.
I can't use abort()
Does this idea violate some HTTP principle?
Thanks

It sounds like you need a "catch-all" endpoint. Typically, it seems a catch-all endpoint would return a generic 404, but in your case, you probably want to return a 200 with some contextual information. Here's basically how you can do it (credit goes to http://flask.pocoo.org/snippets/57/):
from flask import Flask
app = Flask(__name__)
#app.route('/', defaults={'path': ''})
#app.route('/<path:path>')
def catch_all(path):
# returns a 200 (not a 404) with the following contents:
return 'your custom error content\n'
# This is just one of your other valid routes:
#app.route('/stuff')
def stuff():
return 'stuff\n'
if __name__ == '__main__':
app.run()
If you run this and curl various endpoints of the test app, here's what you get:
$ curl localhost:5000/stuff
stuff
$ curl localhost:5000/foo/bar
your custom error content
$ curl localhost:5000/otherstuff
your custom error content
As you can see, your other routes will still work as you expect.

I've decided a solution to this is too hard! I can not find any way to get the browser to ignore a response. There is no response header for 'do nothing'. If there was we would probably never see a webserver error again, which would not be good.
I could ajaxify all the requests as a way to grab the response headers and analyze them before any rendering or redirecting happens. That starts to break all the navigation (back buttons at least) and the pretty URLs. I could bung in a JS routing framework etc, and while I'm leaning how it works I'm not building my app (I already have enough to learn!)
#app.errorhandler(404)
def page_not_found(error):
return redirect(url_for('index'))
If you come up with something great post it anyway, I'm not the first to ask this question, and probably not the last.
Thanks

I remember reading about a javascript library some days ago (but I don't remember the name...). The clou with this library was, that all links and form submits were loaded not directly into the browser "_top" frame/window but into a hidden div and afterwards, when done, the content of the page was replaced by the content of this hidden div.
So if you want to catch bad links and such on client side you could hook up all links and submits and check the http response code. If it is not 200 (ok) you display an error. If it is okay you decide, if you replace the old page with the new content.
But there are two problems with this solution:
1. You would have to change the browsers location (in the address bar) without reloading the page of course!
2. It might get tricky to post some file uploads with javascript.
If I find the link or name of the js-library I saw, I will tell you!