Python 请求响应 404

Question

我可以访问这个网站，我从另一个网站https://www.betexplorer.com/soccer/algeria/ligue-1/bordj-bou-arreridj-tlemcen/tfvAHu7U/，在开发者工具的网络部分。

代码

import requests

url = 'https://www.betexplorer.com/archive-odds/4urejxv464x0xd4645/18/'

response = requests.get(url, headers={'User-Agent': 'my user agent'})

print(response)

输出<Response [404]>

Answer 1

设置Referer HTTP header以获得正确的响应：

import json
import requests


url = "https://www.betexplorer.com/archive-odds/4urejxv464x0xd4645/18/"
headers = {
    "Referer": "https://www.betexplorer.com",
}
data = requests.get(url, headers=headers).json()

print(json.dumps(data, indent=4))

打印：

[
    {
        "date": "23.07. 17:36",
        "odd": "1.88",
        "change": "+0.08"
    },
    {
        "date": "23.07. 17:34",
        "odd": "1.80",
        "change": "-0.01"
    },

...

Python 请求响应 404

Python Requests Response 404

python

request

web-scraping