How to correctly get current URL Address in Python?

Question

I'm running my project on localhost(http://127.0.0.1:5000/) and would be differently on my client server. So I try tried to dynamically get the URL Address but with no avail.

I tried: request.host_url as showing in this thread. However, it returned

RuntimeError: working outside of request context

Here what I tried to do is run python unittest that need full URL as parameter post(), that's a reason I need it dynamic url.

import json, requests

from flask import request
from .base import *

class TestProvince(BaseTest):

    # Test Data
    ID               = '41'
    DESCRIPTION      = 'Province Test'
    LOCAL_DESCRIPTION = 'Province local Description'

    def test_add_record(self):

        print "Adding a new record"

        url = "http://127.0.0.1:5000/Project/API/Province/" # <-- Here I want it to be dynamic url
        payload = {
                    "Status":"AUTH",
                    "Curr":"0",
                    "Inputter":"System",
                    "Createdon":"2019-05-03",
                    "Authorizer":"System",
                    "Authorizeon":"2019-05-03",
                    "Branch":"HO",
                    "ID":self.ID,
                    "Description":self.DESCRIPTION,
                    "LocalDescription":self.LOCAL_DESCRIPTION
                }
        headers = {
            'Content-Type': "application/json",
            'Accept': "application/json"
            }

        response = requests.request("POST", url, data=json.dumps(payload), headers=headers)


if __name__ == "__main__":
    unittest.main()

Any tip for how can I get current URL Address correctly in python? Thanks

@Shadowfax, I have edited question adding code snippet for you already. Just I think it is not code error, that's why I did not post long snippet. Please kindly help. Thanks — Houy Narun, Jul 03 '19 at 09:54

score 0 · Answer 1 · answered Jul 03 '19 at 10:10

0

From request documentation

request.base_url should be what you should be using. I can't find documentation for request.host_url anywhere.

answered Jul 03 '19 at 10:10

deezpeeps

114
2
10

thanks I still get error: working outside of request context. – Houy Narun Jul 03 '19 at 10:17
@HouyNarun from [here](https://stackoverflow.com/a/9932189/3632345), Wrap your thread code in a `test_request_context` – deezpeeps Jul 03 '19 at 10:40

score 0 · Answer 2 · answered Jul 03 '19 at 11:11

The request object has a .host attribute, it will give you the IP and the port where your code is running.

request.host

To get just the host name or IP address, try:

request.host.split(':')[0]

Or you could query the WSGI environment directly:

request.environ['SERVER_NAME']

score 0 · Answer 3 · answered Feb 25 '21 at 01:16

0

From what I've learnt, request.host do not work outside the routes function in flask. For example, do something like this

@app.route('/')
def get_something():
    print(request.host)

This will work because it is in a route function.

answered Feb 25 '21 at 01:16

Paul Adediran

1
2

How to correctly get current URL Address in Python?

3 Answers3