使用 BeautifulSoup 从范围 class 中提取锚文本

Question

这是我要抓取的html：

<span class="meta-attributes__attr-tags">
<a href="/tags/cinematic" title="cinematic">cinematic</a>, 
<a href="/tags/dissolve" title="dissolve">dissolve</a>,
<a href="/tags/epic" title="epic">epic</a>,
<a href="/tags/fly" title="fly">fly</a>,
</span>

我想获取每个 href 的锚文本：电影、溶解、史诗等。

这是我的代码：

url = urllib2.urlopen("http: example.com")

content = url.read()
soup = BeautifulSoup(content)

links = soup.find_all("span", {"class": "meta-attributes__attr-tags"})
for link in links:
    print link.find_all('a')['href']

如果我使用 "link.find_all" 执行此操作，我会收到错误消息：TypeError：列表索引必须是整数，而不是 str。

但是如果我打印 link.find('a')['href'] 我只会得到第一个。

我怎样才能得到所有这些？

Answer 1

您可以执行以下操作：

from bs4 import BeautifulSoup

content = '''
<span class="meta-attributes__attr-tags">
<a href="/tags/cinematic" title="cinematic">cinematic</a>, 
<a href="/tags/dissolve" title="dissolve">dissolve</a>,
<a href="/tags/epic" title="epic">epic</a>,
<a href="/tags/fly" title="fly">fly</a>,
</span>
'''

soup = BeautifulSoup(content)
spans = soup.find_all("span", {"class": "meta-attributes__attr-tags"})
for span in spans:
    links = span.find_all('a')
    for link in links:
        print link['href']

输出

/tags/cinematic
/tags/dissolve
/tags/epic
/tags/fly

Answer 2

link.find_all('a') returns 带有 bs4 标签的列表。您可能希望通过 href 为每个链接编制索引。所以也许这更接近您的需求：

span = soup.find_all("span", {"class": "meta-attributes__attr-tags"})
for links in span:
    for link in links.find_all('a'):
        print(link['href'])

Answer 3

from bs4 import BeautifulSoup

html = """
<span class="meta-attributes__attr-tags">
<a href="/tags/cinematic" title="cinematic">cinematic</a>, 
<a href="/tags/dissolve" title="dissolve">dissolve</a>,
<a href="/tags/epic" title="epic">epic</a>,
<a href="/tags/fly" title="fly">fly</a>,
</span>
"""

soup = BeautifulSoup(html, "lxml")
spans = soup.find_all("span", {"class": "meta-attributes__attr-tags"})

for span in spans:
    for link in span.find_all('a'):
        print link.text, link['href']

另一种更昂贵的方式可能是：

from bs4 import BeautifulSoup

html = """
<span class="meta-attributes__attr-tags">
<a href="/tags/cinematic" title="cinematic">cinematic</a>,
<a href="/tags/dissolve" title="dissolve">dissolve</a>,
<a href="/tags/epic" title="epic">epic</a>,
<a href="/tags/fly" title="fly">fly</a>,
</span>
"""

soup = BeautifulSoup(html, "lxml")
links = soup.find_all("a")

for link in links:
    if 'meta-attributes__attr-tags' not in link.parent.get('class', []):
        continue

    print link.text, link['href']

Answer 4

您可以通过使用 CSS selector:

避免嵌套循环或循环内的任何附加 if 检查

for link in soup.select(".meta-attributes__attr-tags a[href]"):
    print(link["href"], link.get_text())

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