Question

我使用以下代码获取项目的网址：

node.xpath('//td/a[starts-with(text(),"itunes")]')[0].attrib['href']

它给了我类似的东西：

itunes20170107.tbz

但是，我希望获得完整的网址，即：

https://feeds.itunes.apple.com/feeds/epf/v3/full/20170105/incremental/current/itunes20170109.tbz

有没有一种简单的方法可以从lxml获取完整的url，而无需自己构建它？

Answer 1

lxml.html只会解析HTML中的href。如果您想要建立绝对链接而不是相对链接，则应使用urljoin()：

from urllib.parse import urljoin  # Python3
# from urlparse import urljoin  # Python2 

url = "https://feeds.itunes.apple.com/feeds/epf/v3/full/20170105/incremental/current"

relative_url = node.xpath('//td/a[starts-with(text(),"itunes")]')[0].attrib['href']
absolute_url = urljoin(url, relative_url)

演示：

>>> from urllib.parse import urljoin  # Python3
>>> 
>>> url = "https://feeds.itunes.apple.com/feeds/epf/v3/full/20170105/incremental/current"
>>> 
>>> relative_url = "itunes20170107.tbz"
>>> absolute_url = urljoin(url, relative_url)
>>> absolute_url
'https://feeds.itunes.apple.com/feeds/epf/v3/full/20170105/incremental/itunes20170107.tbz'

Answer 2

另一种方法：

import requests
from lxml import fromstring

url = 'http://server.com'
response = reqests.get(url)
etree = fromstring(response.text)
etree.make_links_absolute(url)`

如何从xpath获取绝对URL？

2 个答案: