如何从FTP下载文件? [英] How to download file from ftp?
问题描述
如何从python下载ftp文件?
操作系统 - Windows XP - 如果有区别的话。
这是我目前使用的代码片段
$ b
import mimetypes
import os
import urllib2
import urlparse
def filename_from_url(url):
返回os.path.basename(urlparse.urlsplit(url)[2])
def download_file(url):
创建一个urllib2请求并返回请求加上一些有用的信息
name = filename_from_url(url)
r = urllib2.urlopen(urllib2.Request(url))
info = r。 info()
如果信息中的Content-Disposition:
#如果响应具有Content-Disposition,我们从它取文件名
name = info ['Content-Disposition'] .split ('filename =')[1]
if name [0] ==''or name [0] ==':
name = name [1:-1]
elif r.geturl()!= url:
#如果我们被重定向了,从最后一个网址取文件名
name = filename_from_url(r。 geturl())
content_type = None
如果信息中的'Content-Type':
content_type = info ['Content-Type']。split(';')[0]
#尝试猜测丢失的信息
如果不是名称而不是content_type:
name ='unknown'
elif不是名称:
name ='unknown'+ mimetypes.guess_extension content_type)或''
elif not content_type:
content_type = mimetypes.guess_type(name)[0]
return r,name,content_type
用法:
req,filename,content_type = download_file然后你可以使用 req
作为文件类对象,例如使用 shutil.copyfileobj()
将文件内容复制到本地文件中。如果MIME类型无关紧要,只需删除那部分代码即可。
由于您似乎懒惰,因此代码直接将文件下载到本地文件: / p>
导入shutil
def download_file_locally(url,dest):
req,filename,content_type = download_file )
如果dest.endswith('/'):
dest = os.path.join(dest,filename)
打开(dest,'wb')为f:
shutil.copyfileobj(req,f)
req.close()
如果您指定以斜杠结尾的路径,则足够聪明地使用由服务器发送的文件名,否则使用您指定的目的地。
I'm scripting an install script in python.
How do I download file from ftp in python?
Operating system -- Windows XP - if that makes a difference.
解决方案 Here's a code snippet I'm currently using.
import mimetypes
import os
import urllib2
import urlparse
def filename_from_url(url):
return os.path.basename(urlparse.urlsplit(url)[2])
def download_file(url):
"""Create an urllib2 request and return the request plus some useful info"""
name = filename_from_url(url)
r = urllib2.urlopen(urllib2.Request(url))
info = r.info()
if 'Content-Disposition' in info:
# If the response has Content-Disposition, we take filename from it
name = info['Content-Disposition'].split('filename=')[1]
if name[0] == '"' or name[0] == "'":
name = name[1:-1]
elif r.geturl() != url:
# if we were redirected, take the filename from the final url
name = filename_from_url(r.geturl())
content_type = None
if 'Content-Type' in info:
content_type = info['Content-Type'].split(';')[0]
# Try to guess missing info
if not name and not content_type:
name = 'unknown'
elif not name:
name = 'unknown' + mimetypes.guess_extension(content_type) or ''
elif not content_type:
content_type = mimetypes.guess_type(name)[0]
return r, name, content_type
Usage:
req, filename, content_type = download_file('http://some.url')
Then you can use req
as a file-like object and e.g. use shutil.copyfileobj()
to copy the file contents into a local file. If the MIME type doesn't matter simply remove that part of the code.
Since you seem to be lazy, here's code downloading the file directly to a local file:
import shutil
def download_file_locally(url, dest):
req, filename, content_type = download_file(url)
if dest.endswith('/'):
dest = os.path.join(dest, filename)
with open(dest, 'wb') as f:
shutil.copyfileobj(req, f)
req.close()
This method is smart enough to use the filename sent by the server if you specify a path ending with a slash, otherwise it uses the destination you specified.
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