将JSON对象传递给包含请求的URL

Question

所以,我想使用Kenneth的优秀请求模块.在尝试使用Freebase API时偶然发现了这个问题.

基本上,他们的API看起来像这样:

https://www.googleapis.com/freebase/v1/mqlread?query=...

作为一个查询,他们期望一个JSON对象,这里将返回一个葡萄酒列表及其国家和酒精百分比:

[{
  "country":       null,
  "name":          null,
  "percentage_alcohol": null,
  "percentage_alcohol>": 0,
  "type":          "/food/wine"
}]?

当然,在将它传递给URL之前,我们必须逃避这一点,所以实际的查询将如下所示:

 fullurl = 'https://www.googleapis.com/freebase/v1/mqlread?query=%5B%7B%22percentage_alcohol%3E%22%3A+0%2C+%22country%22%3A+null%2C+%22type%22%3A+%22%2Ffood%2Fwine%22%2C+%22name%22%3A+null%2C+%22percentage_alcohol%22%3A+null%7D%5D'

现在,

r = requests.get(fullurl)
print r.status_code
>>> 400

因为该网站声称它无法解析查询.

r2 = urllib2.urlopen(fullurl)
print r2.getcode()
>>> 200

没问题,我得到了适当的回报.有趣的是,

# This is the url of our requests.get request
print urllib2.urlopen(r.url).getcode() 
>>> 200

为什么？我使用模块错了吗？或者它是一个错误requests？

Answer 1

这个对我有用.这是我做的:

>>> params = [{"country": None,
...            "name": None,
...            "percentage_alcohol": None,
...            "percentage_alcohol>": 0,
...            "type": "/food/wine"
...          }]
>>> import json
>>> params_json = json.dumps(params)

>>> import requests
>>> url = "https://www.googleapis.com/freebase/v1/mqlread?query=%s"
>>> r = requests.get(url % params_json)
>>> r.status_code
200

>>> content_json = json.loads(r.content)
>>> import pprint
>>> pprint.pprint(content_json)
{u'result': [{u'country': u'New Zealand',
              u'name': u'2003 Cloudy Bay Sauvignon Blanc',
              u'percentage_alcohol': 13.5,
              u'type': u'/food/wine'},
             {u'country': u'France',
              u'name': u'G.H. Mumm Cordon Rouge Brut',
              u'percentage_alcohol': 12.0,
              u'type': u'/food/wine'},
....

我为了简洁而减少休息时间.有100个结果.requests.__version__ == '0.10.6'