202

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?

davidism
  • 110,080
  • 24
  • 357
  • 317
Dogukan Tufekci
  • 2,678
  • 3
  • 15
  • 20

4 Answers4

315

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.

booshong
  • 748
  • 6
  • 21
icecrime
  • 71,079
  • 13
  • 98
  • 110
  • 6
    I am trying to get `Request.root_url` and as return I get only `` instead of nicely formatted `http://www.example.com/myapplication/`. Or this feature does not work on localhost? – Vadim Mar 28 '16 at 20:16
  • 6
    @Vadim You should use request.root_url, not Request.root_url. – selfboot Jun 06 '16 at 05:22
  • 3
    new to Flask, i didn’t know where request object come from and how it works, here it is: http://flask.pocoo.org/docs/0.12/reqcontext/ – Ulysse BN Jan 27 '17 at 23:20
  • 3
    request.url_root works for me, whereas request.root_url and Request.root_url fail. Therefore, watch for the cap 'R' and url_root versus root_url – zerocog Aug 03 '17 at 20:50
  • 1
    url_root returns `http://www.example.com/` not `http://www.example.com/myapplication/` base_url returns `http://www.example.com/myapplication/` – moto Dec 04 '18 at 21:27
  • my app route is like @app.route('/index//') how can i guest path from request without variables. I expect **/index** in result. Which flask function to use? – user4772933 Nov 28 '19 at 16:11
  • 2
    @Ian `Request` is a class where you interact with them through `request` which is an instance. This answer doesn't talk about the instance only the class. Leaving one to wonder "how I getz instance?" – Peilonrayz Oct 07 '20 at 01:05
  • @Peilonrayz that's a good point! The answer could be more explicit with how to use/reference those fields, though arguably the linked documentation helps with that. (and to be super pedantic, `request` is actually a proxy). But you're right; For the average user an explicit reference to plain `request.path` etc would be easier to digest. – booshong Oct 07 '20 at 03:02
291

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
chason
  • 3,064
  • 1
  • 6
  • 6
  • 16
    This should be the accepted answer, as it has a lot more detail to it. – pfabri Jun 14 '19 at 14:30
  • 3
    This is a great answer. Very useful and complete. – Tao Starbow Jun 28 '19 at 23:07
  • For people who want to use `request.full_path`, suggest to use `request.environ['RAW_URI']` instead. Because, when the actual full query path is `/alert/dingding/test`, `request.full_path` returns `/alert/dingding/test?`, an extra question mark will be added to the result, which might not be desirable. – AnnieFromTaiwan Nov 24 '20 at 03:59
  • **request.remote_addr** for 127.0.0.1 – PYK Jan 12 '22 at 01:27
12

you should try:

request.url

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

Wai Ha Lee
  • 8,173
  • 68
  • 59
  • 86
Ran
  • 145
  • 1
  • 4
1

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')
Antonio
  • 177
  • 1
  • 11