带有python请求库的zomato api请求

时间:2015-12-02 09:07:16

标签: python json python-requests pycurl zomato-api

Zomato是最受欢迎的餐厅搜索引擎之一,提供免费的api服务......

如果在api请求中使用curl,则效果很好;

curl -X GET --header "Accept: application/json" --header "user_key: MY_API_KEY_HERE" "https://developers.zomato.com/api/v2.1/geocode?lat=41.10867962215988&lon=29.01834726333618"

但是使用了Python的requests库,它不起作用。当我执行下面的代码时;

import requests
r = requests.get("https://developers.zomato.com/api/v2.1/geocode?lat=41.10867962215988&lon=29.01834726333618", headers={"user_key": "MY_API_KEY_HERE", "Accept": "application/json"});

interpreter返回以下错误;

requests.exceptions.ProxyError: Cannot connect to proxy. Socket error: Tunnel connection failed: 403 Forbidden.

通过pyCurl库进行了多次尝试但不幸的结果是相同的; 403 Forbidden

我该如何解决这个问题?

1 个答案:

答案 0 :(得分:8)

我在使用Zomato API时也遇到了问题。我得到了500 Server Error

在标题中添加User Agent信息解决了我的问题。

import requests
from pprint import pprint

locationUrlFromLatLong = "https://developers.zomato.com/api/v2.1/cities?lat=28&lon=77"
header = {"User-agent": "curl/7.43.0", "Accept": "application/json", "user_key": "YOUR_API_USER_KEY"}

response = requests.get(locationUrlFromLatLong, headers=header)

pprint(response.json())