如何处理python中的json数据？

Question

# This code searches the query from googlecustomsearch api and returns data in json format   
import pprint
import json

from googleapiclient.discovery import build


my_api_key = "**************************************"
my_cse_id = "*************************************"

def google_search(search_term, api_key, cse_id, **kwargs):
    service = build("customsearch", "v1", developerKey=api_key)
    res = service.cse().list(q=search_term, cx=cse_id, **kwargs).execute()
    return res['items']


results = google_search(
    'Roshan Patel', my_api_key, my_cse_id, num=10)
for result in results:
    pprint.pprint(result)
    result_dict = json.loads(result)
    print result_dict['formattedUrl']

这是我得到的输出：

我只想获取例如 :u'formattedUrl' 的 url 部分并将其存储在列表中，该怎么做？

新错误

Traceback (most recent call last):

  File "<ipython-input-38-eb898c8de239>", line 1, in <module>
    runfile('C:/Users/abc/untitled9.py', wdir='C:/Users/abc')

  File "C:\Users\abc\Anaconda2\lib\site-packages\spyder\utils\site\sitecustomize.py", line 880, in runfile
    execfile(filename, namespace)

  File "C:\Users\abc\Anaconda2\lib\site-packages\spyder\utils\site\sitecustomize.py", line 87, in execfile
    exec(compile(scripttext, filename, 'exec'), glob, loc)

  File "C:/Users/abc/untitled9.py", line 28, in <module>
    result_dict = json.loads(result)

  File "C:\Users\abc\Anaconda2\lib\json\__init__.py", line 339, in loads
    return _default_decoder.decode(s)

  File "C:\Users\abc\Anaconda2\lib\json\decoder.py", line 364, in decode
    obj, end = self.raw_decode(s, idx=_w(s, 0).end())

TypeError: expected string or buffer

Answer 1

您是否需要先将结果转换为 Python 对象（如字典）才能使用 json 模块：

result_dict = json.loads(result)

现在你可以过滤你的字典了：

result_dict['formattedUrl']

Answer 2

result 不是 JSON，而是一个实际的 Python 字典，所以只是：

print result[u'formattedUrl']

如何处理python中的json数据？

How to deal with json data in python?

python

json

google-custom-search