python—为什么它不能在向api/url发出请求后打印响应？

busg9geu 于 2021-07-14 发布在 Java

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我刚刚通过检查xhr找到了bestbuycaapi。

aboveurl = 'https://www.bestbuy.ca/ecomm-api/availability/products?accept=application%2Fvnd.bestbuy.simpleproduct.v1%2Bjson&accept-language=en-CA&skus=14962185'

我试过了：

response= requests.get(aboveurl)
  print(response.text)

r = requests.get(url).json()
  print(r)

当我在vsc中运行代码时，它会启动并继续运行，但不会显示任何内容。

python Api

来源：https://stackoverflow.com/questions/67281003/why-cant-it-print-response-after-making-a-request-to-an-api-url

1条答案

按热度按时间

kfgdxczn1#

您必须在请求中添加标题才能获得请求结果：

import requests
headers = {'User-Agent': 'Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/56.0.2924.76 Safari/537.36', "Upgrade-Insecure-Requests": "1","DNT": "1","Accept": "text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8","Accept-Language": "en-US,en;q=0.5","Accept-Encoding": "gzip, deflate"}
aboveurl = "https://www.bestbuy.ca/ecomm-api/availability/products?accept=application%2Fvnd.bestbuy.simpleproduct.v1%2Bjson&accept-language=en-CA&skus=14962185"
html = requests.get(aboveurl,headers=headers)
print(f'requrest code: {html.status_code}\n request text: {html.text}')

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python—为什么它不能在向api/url发出请求后打印响应？

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