使用Beautiful Soup在href元素中提取特定的psuedo标记

Question

我有一个锚标记如下：

<a class="gsc_a_at" href= "/citations?view_op=view_citation&amp;hl=en&amp;user=11JgipcAAAAJ&amp;pagesize=100&amp;citation_for_view=11JgipcAAAAJ:j3f4tGmQtD8C">'''

我想使用citation_for_view在beautifulSoup之后提取内容。如果没有regular expressions我该怎么办？

以下是我的尝试。

input_data =＆＃39;＆＃39;＆＃39; ＆＃39;＆＃39;＆＃39;

#!/usr/bin/python
from bs4 import BeautifulSoup

soup = BeautifulSoup(input_data)

for href_tags in soup.find_all('a',href=True):
    print href_tags['href']

输出：

/citations?view_op=view_citation&hl=en&user=11JgipcAAAAJ&pagesize=100&citation_for_view=11JgipcAAAAJ:j3f4tGmQtD8C

如何提取citation_for_view内容href并且仅输出11JgipcAAAAJ:j3f4tGmQtD8C

Answer 1

您可以使用urlparse

>>> import urlparse

>>> url = '/citations?view_op=view_citation&hl=en&user=11JgipcAAAAJ&pagesize=100&citation_for_view=11JgipcAAAAJ:j3f4tGmQtD8C'
>>> vals = urlparse.parse_qs(url)
>>> print vals.get('citation_for_view')
['11JgipcAAAAJ:j3f4tGmQtD8C']