AttributeError : ResultSet object has no attribute 'find_all'

Question:

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whenever i try to get the a tags from the p tags i get this error.
Here’s the code

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)

I have tried this but it still doesn’t fix the error. Any help will be appreciated and thanks in advance.

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

and with this i get the a tags but when i try to get the href i get the same error and and even i get the href somehow it’s only the last href while there should be 19 before it.

p_tags = soup.find_all('p', class_='name')  
for a in p_tags:
    a_tags = a.find_all('a')
    print(len(a_tags))
Asked By: Omega500

||

Answers:

p_tags = soup.find_all('p', class_='name') this return a list. List doesn’t have a find_all() method. iterate over the items in p_tags to call find_all().

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

Your line p_tags = soup.find_all('p', class_='name') returns a result set (e.g., a collection of matching elements). You can’t call find_all on a result set. You need to iterate over the result set.

So you could try something like this:

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)
Answered By: JoshG
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