如何使用BeautifulSoup解析此HTML？

Question

我正在尝试使用Python和BeautifulSoup模块从Acharts.co中获取前100个歌曲排行榜。到目前为止，我已经设法获得了图表中某个特定位置的歌曲标题，但我有点难以获得艺术家的名字

import requests
from bs4 import BeautifulSoup

url = "https://acharts.co/canada_singles_top_100/2021/05"

headers = {
    "Accept": "text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.9", 
    "Accept-Encoding": "gzip, deflate, br", 
    "Accept-Language": "en,de;q=0.9,en-US;q=0.8,fr-FR;q=0.7,fr;q=0.6,es;q=0.5",  
    "authority": "acharts.co", 
    "Upgrade-Insecure-Requests": "1",
    "User-Agent": "Mozilla/5.0 (Windows NT 6.3) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/58.0.3029.110 YaBrowser/17.6.1.749 Yowser/2.5 Safari/537.36"
}    

response = requests.get(url, headers=headers)
response.encoding = 'utf-8'

soup = BeautifulSoup(response.text, 'html.parser')
for item in soup.select("td"):
    if item['class'][0] == 'cPrinciple':
        song = item.a.span.get_text()
        print(song)

下面是我试图解析的HTML部分：

<td class="cPrinciple" itemprop="item" itemscope itemtype="http://schema.org/MusicRecording">
    <a href="https://acharts.co/song/156580" itemprop="url"><span itemprop="name">Mood</span></a>



    <br />
    <span class="Sub">
            <span itemprop="byArtist" itemscope itemtype="http://schema.org/MusicGroup">
                <meta itemprop="url" content="https://acharts.co/artist/24kgoldn" />
                <span itemprop="name">24Kgoldn</span>
            </span> and 
            <span itemprop="byArtist" itemscope itemtype="http://schema.org/MusicGroup">
                <meta itemprop="url" content="https://acharts.co/artist/iann_dior" />
                <span itemprop="name">Iann Dior</span>
            </span>
    </span>

那么在上面的片段中，我将如何提取“情绪”（歌曲标题）、“24kGldn”（艺术家1）和“Iann Dior”（艺术家2）？ 提前谢谢

Answer 1

更简洁的方式（使用列表理解）：

import requests as rq
from bs4 import BeautifulSoup as bs

headers = {"User-Agent": "Mozilla/5.0 (Windows NT 6.3) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/58.0.3029.110 YaBrowser/17.6.1.749 Yowser/2.5 Safari/537.36"}
url = "https://acharts.co/canada_singles_top_100/2021/05"
resp = rq.get(url, headers=headers)
soup = bs(resp.content)

tbody = soup.find_all("tbody")[0]

rows = [[span.text for span in row.find_all("span", attrs={"itemprop": True}) if not "\n" in span.text] for row in tbody.find_all("tr")]

Answer 2

你可以这样做：

soup = BeautifulSoup(response.text, 'html.parser')
for item in soup.select("td"):
    if item['class'][0] == 'cPrinciple':
        e = item.find("span", { "class" : "Sub" })
        if e is not None:
            results= e.find_all("span",{"itemprop":"name"})
            artists = [x.text for x in results]
        song = item.a.span.get_text()
        print(artists)
        print(song)