urllib2 file name

Question

If I open a file using urllib2, like so:

remotefile = urllib2.urlopen('http://example.com/somefile.zip')

Is there an easy way to get the file name other then parsing the original URL?

EDIT: changed openfile to urlopen... not sure how that happened.

EDIT2: I ended up using:

filename = url.split('/')[-1].split('#')[0].split('?')[0]

Unless I'm mistaken, this should strip out all potential queries as well.

Paz

i think it's not corect. I mean urllib2.openfile(...) – Zasuwa

Do make sure you know what you want in these two cases: trailing slash (http://example.com/somefile/) and no path: http://example.com Your example will fail on the latter for sure (returning "example.com"). So will @insin's final answer. That's another reason why using urlsplit is good advice. – Eugenol

from the response headers: #11783769 – Moulder

Lots of answers here miss the fact that there are two places to look for a file name: the URL and the Content-Disposition header field. All the current answers that mention the header neglect to mention that cgi.parse_header() will parse it correctly. There is a better answer here: https://mcmap.net/q/297932/-python-httplib-urllib-get-filename – Descriptive

Answer 1

Did you mean urllib2.urlopen?

You could potentially lift the intended filename if the server was sending a Content-Disposition header by checking remotefile.info()['Content-Disposition'], but as it is I think you'll just have to parse the url.

You could use urlparse.urlsplit, but if you have any URLs like at the second example, you'll end up having to pull the file name out yourself anyway:

>>> urlparse.urlsplit('http://example.com/somefile.zip')
('http', 'example.com', '/somefile.zip', '', '')
>>> urlparse.urlsplit('http://example.com/somedir/somefile.zip')
('http', 'example.com', '/somedir/somefile.zip', '', '')

Might as well just do this:

>>> 'http://example.com/somefile.zip'.split('/')[-1]
'somefile.zip'
>>> 'http://example.com/somedir/somefile.zip'.split('/')[-1]
'somefile.zip'

Volin

Use posixpath.basename() instead of manually splitting on '/'. – Unaccustomed

I would always use urlsplit() and never straight string splitting. The latter will choke if you have an URL that has a fragment or query appended, say example.com/filename.html?cookie=55#Section_3. – Heterophyte

What about escaped characters? Should those be decoded first? – Denney

Answer 2

If you only want the file name itself, assuming that there's no query variables at the end like http://example.com/somedir/somefile.zip?foo=bar then you can use os.path.basename for this:

[user@host]$ python
Python 2.5.1 (r251:54869, Apr 18 2007, 22:08:04) 
Type "help", "copyright", "credits" or "license" for more information.
>>> import os
>>> os.path.basename("http://example.com/somefile.zip")
'somefile.zip'
>>> os.path.basename("http://example.com/somedir/somefile.zip")
'somefile.zip'
>>> os.path.basename("http://example.com/somedir/somefile.zip?foo=bar")
'somefile.zip?foo=bar'

Some other posters mentioned using urlparse, which will work, but you'd still need to strip the leading directory from the file name. If you use os.path.basename() then you don't have to worry about that, since it returns only the final part of the URL or file path.

Knurl

Using os.path to parse URLs seems to rely on current OS splitting paths in the same way as URLs are split. I don't think it's guaranteed for every OS. – Sero

This won't work on Windows. Use import posixpath; posixpath.basename instead. – Albumenize

Answer 3

I think that "the file name" isn't a very well defined concept when it comes to http transfers. The server might (but is not required to) provide one as "content-disposition" header, you can try to get that with remotefile.headers['Content-Disposition']. If this fails, you probably have to parse the URI yourself.

Metastasize

Answer 4

Just saw this I normally do..

filename = url.split("?")[0].split("/")[-1]

Jaundiced

Answer 5

Using urlsplit is the safest option:

url = 'http://example.com/somefile.zip'
urlparse.urlsplit(url).path.split('/')[-1]

Looksee

Answer 6

Do you mean urllib2.urlopen? There is no function called openfile in the urllib2 module.

Anyway, use the urllib2.urlparse functions:

>>> from urllib2 import urlparse
>>> print urlparse.urlsplit('http://example.com/somefile.zip')
('http', 'example.com', '/somefile.zip', '', '')

Voila.

Heterophyte

Answer 7

The os.path.basename function works not only for file paths, but also for urls, so you don't have to manually parse the URL yourself. Also, it's important to note that you should use result.url instead of the original url in order to follow redirect responses:

import os
import urllib2
result = urllib2.urlopen(url)
real_url = urllib2.urlparse.urlparse(result.url)
filename = os.path.basename(real_url.path)

Sigmoid

Answer 8

You could also combine both of the two best-rated answers : Using urllib2.urlparse.urlsplit() to get the path part of the URL, and then os.path.basename for the actual file name.

Full code would be :

>>> remotefile=urllib2.urlopen(url)
>>> try:
>>>   filename=remotefile.info()['Content-Disposition']
>>> except KeyError:
>>>   filename=os.path.basename(urllib2.urlparse.urlsplit(url).path)

Birthright

Answer 9

I guess it depends what you mean by parsing. There is no way to get the filename without parsing the URL, i.e. the remote server doesn't give you a filename. However, you don't have to do much yourself, there's the urlparse module:

In [9]: urlparse.urlparse('http://example.com/somefile.zip')
Out[9]: ('http', 'example.com', '/somefile.zip', '', '', '')

Ube

Answer 10

not that I know of.

but you can parse it easy enough like this:

url = 'http://example.com/somefile.zip'
print url.split('/')[-1]

Urbane

Answer 11

import os,urllib2
resp = urllib2.urlopen('http://www.example.com/index.html')
my_url = resp.geturl()

os.path.split(my_url)[1]

# 'index.html'

This is not openfile, but maybe still helps :)

Zasuwa

Answer 12

using requests, but you can do it easy with urllib(2)

import requests
from urllib import unquote
from urlparse import urlparse

sample = requests.get(url)

if sample.status_code == 200:
    #has_key not work here, and this help avoid problem with names

    if filename == False:

        if 'content-disposition' in sample.headers.keys():
            filename = sample.headers['content-disposition'].split('filename=')[-1].replace('"','').replace(';','')

        else:

            filename = urlparse(sample.url).query.split('/')[-1].split('=')[-1].split('&')[-1]

            if not filename:

                if url.split('/')[-1] != '':
                    filename = sample.url.split('/')[-1].split('=')[-1].split('&')[-1]
                    filename = unquote(filename)

Kandicekandinsky

Answer 13

You probably can use simple regular expression here. Something like:

In [26]: import re
In [27]: pat = re.compile('.+[\/\?#=]([\w-]+\.[\w-]+(?:\.[\w-]+)?$)')
In [28]: test_set 

['http://www.google.com/a341.tar.gz',
 'http://www.google.com/a341.gz',
 'http://www.google.com/asdasd/aadssd.gz',
 'http://www.google.com/asdasd?aadssd.gz',
 'http://www.google.com/asdasd#blah.gz',
 'http://www.google.com/asdasd?filename=xxxbl.gz']

In [30]: for url in test_set:
   ....:     match = pat.match(url)
   ....:     if match and match.groups():
   ....:         print(match.groups()[0])
   ....:         

a341.tar.gz
a341.gz
aadssd.gz
aadssd.gz
blah.gz
xxxbl.gz

Lifeanddeath

Answer 14

Using PurePosixPath which is not operating system—dependent and handles urls gracefully is the pythonic solution:

>>> from pathlib import PurePosixPath
>>> path = PurePosixPath('http://example.com/somefile.zip')
>>> path.name
'somefile.zip'
>>> path = PurePosixPath('http://example.com/nested/somefile.zip')
>>> path.name
'somefile.zip'

Notice how there is no network traffic here or anything (i.e. those urls don't go anywhere) - just using standard parsing rules.

Bombazine