使用Python Requests模块无法访问https://99acres.com的问题求助

99acres.com存在反爬机制，仅添加User-Agent不足以绕过检测，以下是几种可行的解决方法：

方法1：补充完整请求头并设置超时 #

模拟更接近浏览器的请求头，同时避免无限等待：

import requests

url = 'https://99acres.com'
headers = {
    'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:121.0) Gecko/20100101 Firefox/121.0',
    'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,*/*;q=0.8',
    'Accept-Language': 'en-US,en;q=0.5',
    'Accept-Encoding': 'gzip, deflate, br',
    'DNT': '1',
    'Connection': 'keep-alive',
    'Upgrade-Insecure-Requests': '1',
    'Sec-Fetch-Dest': 'document',
    'Sec-Fetch-Mode': 'navigate',
    'Sec-Fetch-Site': 'none',
    'Sec-Fetch-User': '?1'
}

try:
    response = requests.get(url, headers=headers, timeout=10, allow_redirects=True)
    response.raise_for_status()
    print("请求成功，状态码:", response.status_code)
    print(response.text[:500])  # 打印前500字符验证
except requests.exceptions.RequestException as e:
    print("请求失败:", str(e))

方法2：改用支持HTTP/2的库 #

99acres可能优先支持HTTP/2协议，而Requests默认使用HTTP/1.1，可改用httpx库（需先安装：pip install httpx[http2]）：

import httpx

url = 'https://99acres.com'
headers = {
    'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:121.0) Gecko/20100101 Firefox/121.0',
    'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,*/*;q=0.8',
    'Accept-Language': 'en-US,en;q=0.5',
    'Upgrade-Insecure-Requests': '1'
}

with httpx.Client(http2=True, timeout=10) as client:
    response = client.get(url, headers=headers)
    print("请求成功，状态码:", response.status_code)
    print(response.text[:500])

方法3：添加代理IP（若IP被封禁） #

如果本地IP被网站限制，可配置代理：

import requests

url = 'https://99acres.com'
headers = {
    'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:121.0) Gecko/20100101 Firefox/121.0',
    # 其他请求头同方法1
}

# 替换为可用的代理地址
proxies = {
    'http': 'http://your-proxy-address:port',
    'https': 'http://your-proxy-address:port'
}

try:
    response = requests.get(url, headers=headers, timeout=10, proxies=proxies)
    response.raise_for_status()
    print("请求成功")
except requests.exceptions.RequestException as e:
    print("请求失败:", str(e))

方法4：复用浏览器Cookies #

从浏览器开发者工具中复制当前Cookies，添加到请求头中：

headers['Cookie'] = '复制的浏览器Cookies字符串'

内容的提问来源于stack exchange，提问作者Manjushri