How to get pdf filename with Python requests?

Question

I'm using the Python requests library to get a PDF file from the web. This works fine, but I now also want the original filename. If I go to a PDF file in Firefox and click download it already has a filename defined to save the pdf. How do I get this filename?

For example:

import requests
r = requests.get('http://www.researchgate.net/profile/M_Gotic/publication/260197848_Mater_Sci_Eng_B47_%281997%29_33/links/0c9605301e48beda0f000000.pdf')
print r.headers['content-type']  # prints 'application/pdf'

I checked the r.headers for anything interesting, but there's no filename in there. I was actually hoping for something like r.filename..

Does anybody know how I can get the filename of a downloaded PDF file with the requests library?

Bonnibelle

Interesting – I was going to say, "well obviously 0c9605301e48beda0f000000.pdf" (as that is in the request) but fortunately I decided to test it first. And FireFox wants to save it as "Mater Sci Eng B47 (1997) 33.pdf". – Osuna

How are you checking the headers? The filename is there, content-disposition : inline; filename="Mater Sci Eng B47 (1997) 33.pdf". FWIW, many PDFs have a Title embedded in them, but not all, and it may not be easy to access if the PDF is in binary form. – Hooey

Answer 1

It is specified in an http header content-disposition. So to extract the name you would do:

import re
d = r.headers['content-disposition']
fname = re.findall("filename=(.+)", d)[0]

Name extracted from the string via regular expression (re module).

Oman

This wouldn't work if the file name is encoded as utf8. Any suggestion there? – Same

findall returns a list of matches. You would need an index like this fname = re.findall("filename=(.+)", d)[0]. – Cattima

This one is incomplete, a filename can we enclosed in quotes. – Chasten

@Chasten try using "filename=\"(.+)\"" to remove quotes – Happygolucky

Just a side case that sometimes expected filenames are not provided within headers, especially with social media CDN links. In that case, you can formulate your own base name (maybe parse the url for the root filename that you would like to use), and then ascertain the correct extension to use as a suffix with something like resp.headers['Content-Type'].split('/')[-1]. – Sprightly

In my case, the regex did not work because my 'content-disposition' also contains filename=*UTF-8: 'Content-Disposition': "attachment; filename=NameOfTheFile.zip; filename*=UTF-8''NameOfTheFile.zip" – Barkeeper

You can use cgi.parse_header and email.header.decode_header to parse the file name properly – Footloose

@Barkeeper @tony-abou-assaleh, I use unquote(header.split("filename*=")[1].replace('UTF-8\'\'',"")) for Unicode – Mustache

Answer 2

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.

Cattima

I suggest calling urllib.parse.unquote in the else clause so you don't get %20s in the filename. – Bozeman

Answer 3

Apparently, for this particular resource it is in:

r.headers['content-disposition']

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

Drool

Not all responses contain the 'content-disposition' header, but as per one of the comments, it seems they are available in this case. – Sweater

Answer 4

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])

Emilieemiline

Be careful in case there is no "Content-Disposition" header! – Cant

could use something like response.headers.get("Content-Disposition","filename=output.bin") to cover the missing header. – Tabshey

You have to also remove " because it is filename="xxxx.xxx" – Chamorro

Answer 5

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'})

Thorlay

This is the most robust option as it removes optional quotes. – Residual

Answer 6

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.

Trossachs

Answer 7

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 since 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)

Ineffable

Answer 8

Using Python's standard library:

from email.message import EmailMessage

msg = EmailMessage()
msg["Content-Disposition"] = response.headers.get("Content-Disposition")
filename = msg.get_filename()

Like others said, the file name is in the "Content-Disposition" header.

The cgi standard library module used to be the way to parse it, but it's deprecated since py311.

The currently recommended way of parsing is using the email module, which is also part of the standard library.

References:

• https://docs.python.org/3/library/cgi.html
• https://peps.python.org/pep-0594/#cgi
• https://docs.python.org/3/library/email.message.html#email.message.EmailMessage.get_filename

Slipper

This is nice, but EmailMessage from Python 3.12 does not implement RFC 5987 correctly. Setting msg["Content-Disposition"] to "attachment; filename* = utf-8''example.csv" (valid according to the RFC) strips everything after the ;. – Hurlyburly

Answer 9

For me (with requests 2.31) worked with lower caps and with the code bellow:

import request

response = requests.get(file_url)
content_disposition = response.headers.get("content-disposition")
file_name = content_disposition.split("=")[1]

Huoh