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

Answer 1

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.

Answer 2

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', '', '', '')

Answer 3

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)

Answer 4

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.

Answer 5

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.

Answer 6

Using urlsplit is the safest option:

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