Python 3.x 无法从HTML BeautifulSoup对象提取URL

我希望从名为url的html BeautifulSoup对象中提取以下url，如下所示：

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[]试试这个。下面是一个例子：

find_all已经为您提供了一个元素列表；你只需要从每个人那里获得href

from bs4 import BeautifulSoup
import requests

url = 'https://mania.bg/p/pulover-alexander-mcqueen-p409648'

r = requests.get(url)
soup = BeautifulSoup(r.content, 'html.parser')

for a in soup.find_all(
        'a',
        attrs={'class':
               'product sellout product-sellout float-left status-1'}):
    print(a['data-producturl'])

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正确，我正在编辑这个问题，因为这个代码soup.find_all'a'，attrs={'class'：'product sellout product sellout float left status-1}实际上返回列出的html。我需要从中提取url。根据新的要求编辑我的答案。如果仍然没有提供所需的数据，那么您需要定义所需的url。代码现在看起来像这样，工作起来像一个符咒：url=csoup。find_all'a'，attrs={'class'：'product sellout product sellout float left status-1'，href=True for num in RangeURL:url=url[num-1]['href']谢谢你的回答，为了避免添加另一个包，我真的会撒谎。另一个答案我已经找到了，没关系^---^

[('https://mania.bg/p/pulover-alexander-mcqueen-p409648', 'https://mania.bg/p/pulover-alexander-mcqueen-p409648')]

from bs4 import BeautifulSoup
import requests

url = 'https://mania.bg/p/pulover-alexander-mcqueen-p409648'

r = requests.get(url)
soup = BeautifulSoup(r.content, 'html.parser')

for a in soup.find_all(
        'a',
        attrs={'class':
               'product sellout product-sellout float-left status-1'}):
    print(a['data-producturl'])

import requests
import bs4

url = 'https://mania.bg/p/pulover-alexander-mcqueen-p409648'
data = requests.get(url)

soup = bs4.BeautifulSoup(data.text,'html.parser')

urls = soup.find_all('a', attrs={'class': 'product sellout product-sellout float-left status-1'})
for num in range(len(urls)):
    url = urls[num]['href']
    print(url)