通过url获取json数据并在python中使用（simplejson）

Question

我想这必须有一个简单的答案，但我正在努力：我想要一个url（输出json）并在python中的可用字典中获取数据。我陷入了最后一步。

>>> import urllib2
>>> import simplejson
>>> req = urllib2.Request("http://vimeo.com/api/v2/video/38356.json", None, {'user-agent':'syncstream/vimeo'})
>>> opener = urllib2.build_opener()
>>> f = opener.open(req)
>>> f.read()             # this works
'[{"id":"38356","title":"Forgetfulness - Billy Collins Animated Poetry","description":"US Poet Laureate Billy Collins reads his poem ","url":"http:\\/\\/vimeo.com\\/38356","upload_date":"2006-01-24 15:21:03","thumbnail_small":"http:\\/\\/80.media.vimeo.com\\/d1\\/5\\/47\\/74\\/thumbnail-4774968.jpg","thumbnail_medium":"http:\\/\\/80.media.vimeo.com\\/d1\\/5\\/46\\/85\\/thumbnail-4685118.jpg","thumbnail_large":"http:\\/\\/images.vimeo.com\\/87\\/39\\/873998\\/873998_640x480.jpg","user_name":"smjwt","user_url":"http:\\/\\/vimeo.com\\/smjwt","user_portrait_small":"http:\\/\\/bitcast.vimeo.com\\/vimeo\\/portraits\\/defaults\\/d.30.jpg","user_portrait_medium":"http:\\/\\/bitcast.vimeo.com\\/vimeo\\/portraits\\/defaults\\/d.75.jpg","user_portrait_large":"http:\\/\\/bitcast.vimeo.com\\/vimeo\\/portraits\\/defaults\\/d.100.jpg","user_portrait_huge":"http:\\/\\/bitcast.vimeo.com\\/vimeo\\/portraits\\/defaults\\/d.300.jpg","stats_number_of_likes":"281","stats_number_of_plays":"9173","stats_number_of_comments":23,"duration":"112","width":"320","height":"240","tags":"poetry, poet, online poetry, famous poet, video poetry, modern poetry, famous poem, poetry sites, poetry websites, audio poetry, american poet, animation clips, american poetry, free poetry sites, animation art, free poetry, animated clips, poem, poet laureate"}]'
>>> simplejson.load(f)
Traceback (most recent call last):
  File "<console>", line 1, in <module>
  File "/usr/lib/python2.5/site-packages/django/utils/simplejson/__init__.py", line 298, in load
    parse_constant=parse_constant, **kw)
  File "/usr/lib/python2.5/site-packages/django/utils/simplejson/__init__.py", line 338, in loads
    return _default_decoder.decode(s)
  File "/usr/lib/python2.5/site-packages/django/utils/simplejson/decoder.py", line 326, in decode
    obj, end = self.raw_decode(s, idx=_w(s, 0).end())
  File "/usr/lib/python2.5/site-packages/django/utils/simplejson/decoder.py", line 344, in raw_decode
    raise ValueError("No JSON object could be decoded")
ValueError: No JSON object could be decoded

我出错的任何想法？

Answer 1

尝试

f = opener.open(req)
simplejson.load(f)

首先不运行f.read（）。当你运行f.read（）时，文件句柄的内容会变得混乱，所以当你的调用 simplejson.load（f）

时，什么都没有了。

Answer 2

第一行读取整个文件。然后第二行尝试从文件中读取更多内容，但没有任何内容：

>>> f.read()             # this works
blah blah blah
>>> simplejson.load(f)

要么省略f.read（）行，要么保存read值，并在加载中使用它：

json = f.read()
simplejson.loads(json)

Answer 3

有一个更简单的方法 - 你根本不需要simplejson。只要将true / false / null设置为正确的值，Python就可以使用eval语句将json解析为dict / array。

# fetch the url
url = "https://api.twitter.com/1/users/lookup.json?user_id=6253282,18949452"
json = urllib2.urlopen(url).read()

# convert to a native python object
(true,false,null) = (True,False,None)
profiles = eval(json)