ResultSet object has no attribute 'find_all'

Question

i always met one problem, when I scraping one web page.

AttributeError: ResultSet object has no attribute 'find'. You're probably treating a list of items like a single item. Did you call find_all() when you meant to call find()?

anyone can tell me how to solve this? my code as below:

import requests  
r = requests.get('https://www.example.com')
from bs4 import BeautifulSoup  
soup = BeautifulSoup(r.text, 'html.parser')  
results = soup.find_all('div', attrs={'class':'product-item item-template-0 alternative'})
records = []  
for result in results:  
    name = results.find('div', attrs={'class':'name'}).text 
    price = results.find('div', attrs={'class':'price'}).text[13:-11]
    records.append((name, price,))

I want to ask a close question.If I want to scrap multiple pages.the pattern like below,I use the code as below,but still scrap the first page only Can you solve this issue.

import requests  
for i in range(100):   
    url = "https://www.example.com/a/a_{}.format(i)"
    r = requests.get(url)
from bs4 import BeautifulSoup  
soup = BeautifulSoup(r.text, 'html.parser')  
results = soup.find_all('div', attrs={'class':'product-item item-template-0 alternative'})

Jacquetta

Does this answer your question? Beautiful Soup: 'ResultSet' object has no attribute 'find_all'? – Catherin

Answer 1

Try this. You mixed up results with result:

import requests  
r = requests.get('https://www.example.com')
from bs4 import BeautifulSoup  
soup = BeautifulSoup(r.text, 'html.parser')  
results = soup.find_all('div', attrs={'class':'product-item item-template-0 alternative'})
records = []  
for result in results:  
    name = result.find('div', attrs={'class':'name'}).text # result not results
    price = result.find('div', attrs={'class':'price'}).text[13:-11]
    records.append((name, price,))

Grazia

thanks, it works, just change the results to be result. – Jacquetta

I want to ask a close question.I want to scrap multiple pages.example.com/a/a_1 example.com/a/a_2 example.com/a/a_3 -------.I use the code as below,but still scrap the first page only Can you solve this issue.import requests for i in range(100): url = "example.com/a/a_{}.format(i)" r = requests.get(url) from bs4 import BeautifulSoup soup = BeautifulSoup(r.text, 'html.parser') results = soup.find_all('div', attrs={'class':'product-item item-template-0 alternative'}) – Jacquetta

You should post another question for this. Why are you looping 100 times? – Grazia

because it has 100 pages – Jacquetta

Oh, I see. Instead of looping 100 times, loop through the list of pages (and index it if necessary). – Grazia

Can you show me the code for reference. high appreciated! – Jacquetta

can you help answer my question?#49106272 – Jacquetta

Answer 2

Try this, remove 's' in 'results' in particularly name = results

your error code "name = results.find('div', attrs={'class':'name'}).text"

with one changes "name = result.find('div', attrs={'class':'name'}).text"

well, nice try!

import requests  
r = requests.get('https://www.example.com')
from bs4 import BeautifulSoup  
soup = BeautifulSoup(r.text, 'html.parser')  
results = soup.find_all('div', attrs={'class':'product-item item-template-0 alternative'})
records = []  
for result in results:  
    name = result.find('div', attrs={'class':'name'}).text 
    price = result.find('div', attrs={'class':'price'}).text[13:-11]
    records.append((name, price,))

Kisner