How to get pdf filename with Python requests?

Apparently, for this particular resource it is in:

r.headers['content-disposition']

Don’t know if it is always the case, though.

Building on some of the other answers, here’s how I do it. If there isn’t a Content-Disposition header, I parse it from the download URL:

import re
import requests
from requests.exceptions import RequestException


url = 'http://www.example.com/downloads/sample.pdf'

try:
    with requests.get(url) as r:

        fname = ''
        if "Content-Disposition" in r.headers.keys():
            fname = re.findall("filename=(.+)", r.headers["Content-Disposition"])[0]
        else:
            fname = url.split("/")[-1]

        print(fname)
except RequestException as e:
    print(e)

There are arguably better ways of parsing the URL string, but for simplicity I didn’t want to involve any more libraries.

You can use werkzeug for options headers https://werkzeug.palletsprojects.com/en/0.15.x/http/#werkzeug.http.parse_options_header

>>> import werkzeug


>>> werkzeug.http.parse_options_header('text/html; charset=utf8')
('text/html', {'charset': 'utf8'})

easy python3 implementation to get filename from Content-Disposition:

import requests
response = requests.get(<your-url>)
print(response.headers.get("Content-Disposition").split("filename=")[1])

According to the documentation, neither Content-Disposition nor its filename attribute is required. Also, I checked dozens links on the internet and haven’t found responses with the Content-Disposition header. So, in most cases, I wouldn’t rely on it much and just retrieve this information from the request URL (note: I’m taking it from req.url because there could be redirection and we want to get real filename). I used werkzeug because it looks more robust and handles quoted and unquoted filenames. Eventually, I came up with this solution (works from Python 3.8):

from urllib.parse import urlparse

import requests
import werkzeug


def get_filename(url: str):
    try:
        with requests.get(url) as req:
            if content_disposition := req.headers.get("Content-Disposition"):
                param, options = werkzeug.http.parse_options_header(content_disposition)
                if param == 'attachment' and (filename := options.get('filename')):
                    return filename

            path = urlparse(req.url).path
            name = path[path.rfind('/') + 1:]
            return name
    except requests.exceptions.RequestException as e:
        raise e

I wrote some tests using pytest and requests_mock:

import pytest
import requests
import requests_mock

from main import get_filename

TEST_URL = 'https://pwrk.us/report.pdf'


@pytest.mark.parametrize(
    'headers,expected_filename',
    [
        (
                {'Content-Disposition': 'attachment; filename="filename.pdf"'},
                "filename.pdf"
        ),
        (
                # The string following filename should always be put into quotes;
                # but, for compatibility reasons, many browsers try to parse unquoted names that contain spaces.
                # https://developer.mozilla.org/en-US/docs/Web/HTTP/Headers/Content-Disposition#directives
                {'Content-Disposition': 'attachment; filename=filename with spaces.pdf'},
                "filename with spaces.pdf"
        ),
        (
                {'Content-Disposition': 'attachment;'},
                "report.pdf"
        ),
        (
                {'Content-Disposition': 'inline;'},
                "report.pdf"
        ),
        (
                {},
                "report.pdf"
        )
    ]
)
def test_get_filename(headers, expected_filename):
    with requests_mock.Mocker() as m:
        m.get(TEST_URL, text='resp', headers=headers)
        assert get_filename(TEST_URL) == expected_filename


def test_get_filename_exception():
    with requests_mock.Mocker() as m:
        m.get(TEST_URL, exc=requests.exceptions.RequestException)
        with pytest.raises(requests.exceptions.RequestException):
            get_filename(TEST_URL)

Use urllib.request instead of requests because then you can do urllib.request.urlopen(…).headers.get_filename(), which is safer than some of the other answers for the following reason:

If the [Content-Disposition] header does not have a filename parameter, this method falls back to looking for the name parameter on the Content-Type header.

After that, even safer would be to additionally fall back to the filename in the URL, as another answer does.