AttributeError: ResultSet object 沒有屬性 'find_all'

Question

這是 html 代碼

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每當我嘗試從 p 標簽獲取 a 標簽時，我都會收到此錯誤。 這是代碼

from urllib import response
from venv import create
from bs4 import BeautifulSoup
import requests

url = "https://gogoanime.gg?page=1"

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


p_tags = soup.find_all('p', class_='name')  
a_tags = p_tags.find_all('a')
for link in a_tags:
    links = link.get('href')
    print(links)

我試過這個但它仍然沒有解決錯誤。 任何幫助將不勝感激並提前致謝。

p_tags = p_tags = soup.find_all('p', class_='name')  
for a_tags in p_tags.find_all('a')
    print(a_tags)

有了這個，我得到了 a 標簽，但是當我嘗試獲取 href 時，我得到了同樣的錯誤，甚至我以某種方式得到了 href，它只是最后一個 href，而它之前應該有 19 個。

p_tags = soup.find_all('p', class_='name')  
for a in p_tags:
    a_tags = a.find_all('a')
    print(len(a_tags))

Answer 1

p_tags = soup.find_all('p', class_='name')這會返回一個列表。 List 沒有find_all()方法。 遍歷p_tags中的項目以調用find_all() 。

for ptag in p_tags:
    a_tags = ptag.find_all('a')
    for link in a_tags:
        links = link.get('href')
        print(links)

Answer 2

您的行p_tags = soup.find_all('p', class_='name')返回一個結果集（例如，匹配元素的集合）。 您不能在結果集上調用find_all 。 您需要遍歷結果集。

所以你可以嘗試這樣的事情：

from urllib import response
from venv import create
from bs4 import BeautifulSoup
import requests

url = "https://gogoanime.gg?page=1"
req = requests.get(url)
Response = req.content
soup = BeautifulSoup(Response, 'html.parser')


p_tags = soup.find_all('p', class_='name')  # returns result set
a_tags = []
for p_tag in p_tags:  # iterate over result set
    a_tags.append(p_tag.find('a', href=True))  # append `a` tag to list

links = []
for a_tag in a_tags:
    links.append(a_tag.get('href'))  # get `href` value for each `a` tag
    
print(links)