How to download a file using python in a 'smarter' way?

I need to download several files via http in Python.

The most obvious way to do it is just using urllib2:

import urllib2
u = urllib2.urlopen('http://server.com/file.html')
localFile = open('file.html', 'w')
localFile.write(u.read())
localFile.close()

But I'll have to deal with the URLs that are nasty in some way, say like this: http://server.com/!Run.aspx/someoddtext/somemore?id=121&m=pdf. When downloaded via the browser, the file has a human-readable name, ie. accounts.pdf.

Is there any way to handle that in python, so I don't need to know the file names and hardcode them into my script?

Download scripts like that tend to push a header telling the user-agent what to name the file:

Content-Disposition: attachment; filename="the filename.ext"

If you can grab that header, you can get the proper filename.

There's another thread that has a little bit of code to offer up for Content-Disposition-grabbing.

remotefile = urllib2.urlopen('http://example.com/somefile.zip')
remotefile.info()['Content-Disposition']

Basierend auf Kommentaren und @ Olis Antwort habe ich eine Lösung wie diese gefunden:

from os.path import basename
from urlparse import urlsplit

def url2name(url):
    return basename(urlsplit(url)[2])

def download(url, localFileName = None):
    localName = url2name(url)
    req = urllib2.Request(url)
    r = urllib2.urlopen(req)
    if r.info().has_key('Content-Disposition'):
        # If the response has Content-Disposition, we take file name from it
        localName = r.info()['Content-Disposition'].split('filename=')[1]
        if localName[0] == '"' or localName[0] == "'":
            localName = localName[1:-1]
    elif r.url != url: 
        # if we were redirected, the real file name we take from the final URL
        localName = url2name(r.url)
    if localFileName: 
        # we can force to save the file as specified name
        localName = localFileName
    f = open(localName, 'wb')
    f.write(r.read())
    f.close()

Es übernimmt den Dateinamen von Content-Disposition; Wenn es nicht vorhanden ist, wird der Dateiname aus der URL verwendet (wenn eine Umleitung erfolgt ist, wird die endgültige URL berücksichtigt).

Hier finden Sie eine pythonischere Lösung:

import urllib2
import shutil
import urlparse
import os

def download(url, fileName=None):
    def getFileName(url,openUrl):
        if 'Content-Disposition' in openUrl.info():
            # If the response has Content-Disposition, try to get filename from it
            cd = dict(map(
                lambda x: x.strip().split('=') if '=' in x else (x.strip(),''),
                openUrl.info()['Content-Disposition'].split(';')))
            if 'filename' in cd:
                filename = cd['filename'].strip("\"'")
                if filename: return filename
        # if no filename was found above, parse it out of the final URL.
        return os.path.basename(urlparse.urlsplit(openUrl.url)[2])

    r = urllib2.urlopen(urllib2.Request(url))
    try:
        fileName = fileName or getFileName(url,r)
        with open(fileName, 'wb') as f:
            shutil.copyfileobj(r,f)
    finally:
        r.close()

2 Kender :

if localName[0] == '"' or localName[0] == "'":
    localName = localName[1:-1]

Es ist nicht sicher - der Webserver kann einen falsch formatierten Namen als ["file.ext] oder [file.ext '] übergeben oder sogar leer sein und localName [0] löst eine Ausnahme aus. Der richtige Code kann folgendermaßen aussehen:

localName = localName.replace('"', '').replace("'", "")
if localName == '':
    localName = SOME_DEFAULT_FILE_NAME

Verwenden von wget:

custom_file_name = "/custom/path/custom_name.ext"
wget.download(url, custom_file_name)

Verwenden von urlretrieve:

urllib.urlretrieve(url, custom_file_name)

urlretrieve erstellt auch die Verzeichnisstruktur, falls nicht vorhanden.