如何为每个<li class=”“><a>找到最近的上述同级<li>？

Question

我的例子是html

<li><h4>A0: Pronouns</h4></li>
<li class="">
    <a>bb</a>
    <a>cc</a>
</li>
<li class="">
    <a>dd</a>
    <a>ee</a>
</li>
<li><h4>A0: Verbs Tenses & Conjugation</h4></li>
<li class="">
    <a>ff</a>
    <a>gg</a>
</li>
<li class="">
    <a>hh</a>
    <a>kk</a>
</li>
<li class="">
    <a>jj</a>
    <a>ii</a>
</li>

对于每个元素<li class=""><a>，我想找到其最近的上面的兄弟元素<li><h4>。比如说,

• <li class=""><a>bb</a></li>对应于<li><h4>A0: Pronouns</h4></li>

• <li class=""><a>dd</a></li>对应于<li><h4>A0: Pronouns</h4></li>

• <li class="">ff<a>dd</a></li>对应于<li><h4>A0: Verbs Tenses & Conjugation</h4></li>

• <li class="">hh<a>dd</a></li>对应于<li><h4>A0: Verbs Tenses & Conjugation</h4></li>

• <li class="">jj<a>dd</a></li>对应于<li><h4>A0: Verbs Tenses & Conjugation</h4></li>

你能详细说明一下怎么做吗

import requests
from bs4 import BeautifulSoup

session = requests.Session()
headers = {
    'Accept-Encoding': 'gzip, deflate, sdch',
    'Accept-Language': 'en-US,en;q=0.8',
    'Upgrade-Insecure-Requests': '1',
    'User-Agent': 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/56.0.2924.87 Safari/537.36',
    'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8',
    'Cache-Control': 'max-age=0',
    'Connection': 'keep-alive',
}

link = 'https://french.kwiziq.com/revision/grammar'
r = session.get(link, headers = headers)
soup = BeautifulSoup(r.content, 'html.parser')

for d in soup.select('.callout-body > ul li > a:nth-of-type(1)'):
    print(d)

Answer 1

您可以使用.find_previous('h4')：

import requests
from bs4 import BeautifulSoup


url = "https://french.kwiziq.com/revision/grammar"

soup = BeautifulSoup(requests.get(url).content, "html.parser")
for a in soup.select(".callout  li > a:nth-of-type(1)"):
    print(
        "{:<70} {}".format(
            a.get_text(strip=True), a.find_previous("h4").get_text(strip=True)
        )
    )

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...

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