Question

问题：当我尝试执行脚本时，BeautifulSoup(html, ...)会给出错误消息＆＃34; TypeError：类型为＆＃39;响应＆＃39;的对象没有len（）。我尝试将实际的html作为参数传递，但它仍然无法正常工作。

import requests

url = 'http://vineoftheday.com/?order_by=rating'
response = requests.get(url)
html = response.content

soup = BeautifulSoup(html, "html.parser")

Answer 1

您将获得"jdbc:mysql:loadbalance://" + hosts + "/test"。但它将响应体返回为字节（docs）。但是你应该将response.content传递给BeautifulSoup构造函数（docs）。因此，您需要使用str而不是获取内容。

Answer 2

尝试直接传递HTML文本

soup = BeautifulSoup(html.text)

Answer 3

html.parser 用于忽略页面中的警告：

soup = BeautifulSoup(html.text, "html.parser")

Answer 4

如果您使用requests.get('https://example.com')获取HTML，则应使用requests.get('https://example.com').text。

Answer 5

您仅在“响应”中获得响应代码并始终使用浏览器标头来确保安全性您将面临许多问题

在调试器控制台网络部分“标头” UserAgent中查找标头

尝试

import requests
from bs4 import BeautifulSoup

from fake_useragent import UserAgent

url = 'http://www.google.com'
headers = {'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_13_6) 
AppleWebKit/537.36 (KHTML, like Gecko) Chrome/71.0.3578.98 Safari/537.36'}

response = requests.get(quote_page, headers=headers).text

soup = BeautifulSoup(response, 'html.parser')
print(soup.prettify())

Answer 6

对我有用：

soup = BeautifulSoup(requests.get("your_url").text)

现在，下面的代码更好（使用lxml解析器）：

import requests
from bs4 import BeautifulSoup

soup = BeautifulSoup(requests.get("your_url").text, 'lxml')

Answer 7

您应该使用 .text 来获取回复内容

import  requests
url = 'http://www ... '
response = requests.get(url)
print(response.text)

或与肥皂

一起使用

import  requests
from bs4 import BeautifulSoup

url = 'http://www ... '
response = requests.get(url)
msg = response.text
print(BeautifulSoup(msg,'html.parser'))

Answer 8

import requests
from urllib.request import urlopen
from bs4 import BeautifulSoup
import re

url = "https://fortnitetracker.com/profile/all/DakshRungta123"
html = requests.get(url)

soup = BeautifulSoup(html)


title = soup.text
print(title.text)

BeautifulSoup：类型对象＆＃39;响应＆＃39;没有len（）

8 个答案: