ResultSet 对象没有属性“find_all”

Question

我在抓取一个网页时总是遇到一个问题。

AttributeError：ResultSet 对象没有“find”属性。您可能将项目列表视为单个项目。当你打算调用 find() 时，你调用了 find_all() 吗？

谁能告诉我如何解决这个问题？我的代码如下：

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,))

我想问一个很接近的问题。如果我想报废多个页面。像下面这样的模式，我使用下面的代码，但仍然只报废第一页你能解决这个问题吗？

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'})

Answer 1

试试这个，删除 'results' 中的 's' 特别是 name = results

您的错误代码“name = results.find('div', attrs={'class':'name'}).text”

一改“name = result.find('div', attrs={'class':'name'}).text”

嗯，不错的尝试！

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,))

Answer 2

试试这个。你把结果和结果混在一起了：

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,))