我试图在Python中开始使用JSON,但似乎我误解了JSON概念中的某些内容。我跟着google api example,这很好。但是当我在JSON响应中将代码更改为较低级别时(如下所示,我尝试访问该位置),我收到以下代码的以下错误消息:
追踪(最近的呼叫最后):
文件“geoCode.py”,第11行,中<module>
test = json.dumps([s ['location'] for s in jsonResponse ['results']], indent = 3)KeyError:'location'
如何在python中访问JSON文件中的较低信息级别?我是否必须转到更高级别并搜索结果字符串?这对我来说似乎很奇怪?
以下是我试图运行的代码:
import urllib, json
URL2 = "http://maps.googleapis.com/maps/api/geocode/json?address=1600+Amphitheatre+Parkway,+Mountain+View,+CA&sensor=false"
googleResponse = urllib.urlopen(URL2);
jsonResponse = json.loads(googleResponse.read())
test = json.dumps([s['location'] for s in jsonResponse['results']], indent=3)
print test
答案 0 :(得分:17)
理解jsonResponse
格式的关键是打印出来:
import urllib, json
import pprint
URL2 = "http://maps.googleapis.com/maps/api/geocode/json?address=1600+Amphitheatre+Parkway,+Mountain+View,+CA&sensor=false"
googleResponse = urllib.urlopen(URL2)
jsonResponse = json.loads(googleResponse.read())
pprint.pprint(jsonResponse)
# {u'results': [{u'address_components': [{u'long_name': u'1600',
# u'short_name': u'1600',
# u'types': [u'street_number']},
# {u'long_name': u'Amphitheatre Pkwy',
# u'short_name': u'Amphitheatre Pkwy',
# u'types': [u'route']},
# {u'long_name': u'Mountain View',
# u'short_name': u'Mountain View',
# u'types': [u'locality',
# u'political']},
# {u'long_name': u'San Jose',
# u'short_name': u'San Jose',
# u'types': [u'administrative_area_level_3',
# u'political']},
# {u'long_name': u'Santa Clara',
# u'short_name': u'Santa Clara',
# u'types': [u'administrative_area_level_2',
# u'political']},
# {u'long_name': u'California',
# u'short_name': u'CA',
# u'types': [u'administrative_area_level_1',
# u'political']},
# {u'long_name': u'United States',
# u'short_name': u'US',
# u'types': [u'country',
# u'political']},
# {u'long_name': u'94043',
# u'short_name': u'94043',
# u'types': [u'postal_code']}],
# u'formatted_address': u'1600 Amphitheatre Pkwy, Mountain View, CA 94043, USA',
# u'geometry': {u'location': {u'lat': 37.4216227,
# u'lng': -122.0840263},
# u'location_type': u'ROOFTOP',
# u'viewport': {u'northeast': {u'lat': 37.424770299999999,
# u'lng': -122.0808787},
# u'southwest': {u'lat': 37.418475100000002,
# u'lng': -122.0871739}}},
# u'types': [u'street_address']}],
# u'status': u'OK'}
test = json.dumps([s['geometry']['location'] for s in jsonResponse['results']], indent=3)
print(test)
# [
# {
# "lat": 37.4216227,
# "lng": -122.0840263
# }
# ]
jsonResponse
是一个词典。jsonResponse['results']
是一个词典列表。for s in jsonResponse['results']
分配
s
到每次迭代的dict
通过循环。s['geometry']
是一个词典。s['geometry']['location']
(最后!)包含
纬度/经度字典。答案 1 :(得分:0)
首先让它感到困惑,导致导入json无法使用加载,它必须是
将simplejson导入为json,所以:
import urllib
import simplejson as json
import pprint
URL2 = "http://pbx/a/kiosks"
googleResponse = urllib.urlopen(URL2)
jsonResponse = json.loads(googleResponse.read())
pprint.pprint(jsonResponse)