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.
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' 
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.
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.
import posixpath; posixpath.basename instead.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.
Just saw this I normally do..
filename = url.split("?")[0].split("/")[-1] Using urlsplit is the safest option:
url = 'http://example.com/somefile.zip' urlparse.urlsplit(url).path.split('/')[-1] 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.
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) 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) 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', '', '', '') not that I know of.
but you can parse it easy enough like this:
url = 'http://example.com/somefile.zip' print url.split('/')[-1]
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) 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 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.
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 :)
http://example.com/somefile/) and no path:http://example.comYour 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.