How do I get the different parts of a Flask request's url?

Question

I want to detect if the request came from the localhost:5000 or foo.herokuapp.com host and what path was requested. How do I get this information about a Flask request?

Asked By: Dogukan Tufekci

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Source

Answer 1

You can examine the url through several Request fields:

Imagine your application is listening on the following application root:
http://www.example.com/myapplication
And a user requests the following URI:
http://www.example.com/myapplication/foo/page.html?x=y
In this case the values of the above mentioned attributes would be the following:
    path             /foo/page.html
    full_path        /foo/page.html?x=y
    script_root      /myapplication
    base_url         http://www.example.com/myapplication/foo/page.html
    url              http://www.example.com/myapplication/foo/page.html?x=y
    url_root         http://www.example.com/myapplication/

You can easily extract the host part with the appropriate splits.

An example of using this:

from flask import request

@app.route('/')
def index():
    return request.base_url

Answered By: icecrime

Answer 2

you should try:

request.url

It suppose to work always, even on localhost (just did it).

Answered By: Ran

Answer 3

another example:

request:

curl -XGET http://127.0.0.1:5000/alert/dingding/test?x=y

then:

request.method:              GET
request.url:                 http://127.0.0.1:5000/alert/dingding/test?x=y
request.base_url:            http://127.0.0.1:5000/alert/dingding/test
request.url_charset:         utf-8
request.url_root:            http://127.0.0.1:5000/
str(request.url_rule):       /alert/dingding/test
request.host_url:            http://127.0.0.1:5000/
request.host:                127.0.0.1:5000
request.script_root:
request.path:                /alert/dingding/test
request.full_path:           /alert/dingding/test?x=y

request.args:                ImmutableMultiDict([('x', 'y')])
request.args.get('x'):       y

Answered By: chason

Answer 4

If you are using Python, I would suggest by exploring the request object:

dir(request)

Since the object support the method dict:

request.__dict__

It can be printed or saved. I use it to log 404 codes in Flask:

@app.errorhandler(404)
def not_found(e):
    with open("./404.csv", "a") as f:
        f.write(f'{datetime.datetime.now()},{request.__dict__}n')
    return send_file('static/images/Darknet-404-Page-Concept.png', mimetype='image/png')

Answered By: Antonio

Answer 5

if user requests the following URI:

http://www.example.com/myapplication/foo/page.html?x=y

and the user wants y

you can use

request.args.get("x")

Answered By: Abdalrhman Qabani

How do I get the different parts of a Flask request's url?

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