在python中存儲這個JSON的正確方法？

Question

我正在嘗試存儲從網站https://www.xkcd.com/info.0.json返回的 JSON 對象

我試過的

url = 'https://www.xkcd.com/info.0.json'
response = requests.get(url)
if response.status_code == 200:
    response_content = str(response.json())
    print(response_content)
    new_response = response_content.replace("'", '"')
    json_data = json.loads(new_response)
    print(new_response)
    print(json_data)

print(response_content)返回

{
    'link': '',
    'month': '11',
    'num': 1603,
    'title': 'Flashlights',
    'safe_title': 'Flashlights',
    'year': '2015',
    'day': '13',
    'img': 'http: //imgs.xkcd.com/comics/flashlights.png',
    'transcript': '',
    'news': '',
    'alt': "Due to a typo, I initially found a forum for serious Fleshlight enthusiasts, and it turns out their highest-end models are ALSO capable of setting trees on fire. They're impossible to use without severe burns, but some of them swear it's worth it."
}

要轉換response_content的單引號，我試過

new_response = response_content.replace("'", '"')

但是問題出現在alt所在的行上

    .....
    "news": "",
    "alt": "Due to a typo, ...... of setting trees on fire. They"reimpossibletousewithoutsevereburns,
butsomeofthemswearit"s worth it.",
}

如果任何值中有單引號，則此方法失敗。

錯誤日志：

 File "./main.py", line 55, in download_latest
    json_data = json.loads(new_response)
  File "/usr/lib/python3.4/json/__init__.py", line 318, in loads
    return _default_decoder.decode(s)
  File "/usr/lib/python3.4/json/decoder.py", line 343, in decode
    obj, end = self.raw_decode(s, idx=_w(s, 0).end())
  File "/usr/lib/python3.4/json/decoder.py", line 359, in raw_decode
    obj, end = self.scan_once(s, idx)
ValueError: Expecting ',' delimiter: line 1 column 342 (char 341)

在腳本中加載 JSON 的任何其他方法？

編輯：

我想做類似的事情

    json_data = json.dumps(response_content)
    print(type(json_data))     ## returns <class 'str'>
    print(json_data['num'])    

但這會返回一個TypeError

File "./main.py", line 53, in download_latest
    print(json_data['num'])
TypeError: string indices must be integers

Answer 1

response.json()方法返回Python 數據結構。 你在這里做了很多事情，你只需要：

url = 'https://www.xkcd.com/info.0.json'
response = requests.get(url)
if response.status_code == 200:
    json_data = response.json()

就是這樣。

您正在將 Python 數據結構轉換為字符串，然后再次嘗試將該字符串解釋為 JSON。 這可能看起來像它的工作str() ，因為 Python 容器的str()轉換使用 Python 語法來生成結果。 但是 Python 不是 JSON，不幸的是，您嘗試將它變成 JSON 也不是很好。 並且根本不需要。

可以直接使用json_data ，它是一個Python字典：

>>> import requests
>>> url = 'https://www.xkcd.com/info.0.json'
>>> response = requests.get(url)
>>> response.status_code
200
>>> json_data = response.json()
>>> type(json_data)
<type 'dict'>
>>> json_data
{u'img': u'http://imgs.xkcd.com/comics/flashlights.png', u'title': u'Flashlights', u'month': u'11', u'num': 1603, u'link': u'', u'year': u'2015', u'news': u'', u'safe_title': u'Flashlights', u'transcript': u'', u'alt': u"Due to a typo, I initially found a forum for serious Fleshlight enthusiasts, and it turns out their highest-end models are ALSO capable of setting trees on fire. They're impossible to use without severe burns, but some of them swear it's worth it.", u'day': u'13'}
>>> print json_data['title']
Flashlights
>>> print json_data['alt']
Due to a typo, I initially found a forum for serious Fleshlight enthusiasts, and it turns out their highest-end models are ALSO capable of setting trees on fire. They're impossible to use without severe burns, but some of them swear it's worth it.

Answer 2

response.json()已經返回了一個 python 字典：

import requests
import json
url = 'https://www.xkcd.com/info.0.json'
response = requests.get(url)
if response.status_code == 200:
    response_content = response.json()
    print response_content

無需在字符串之間進行轉換。

Answer 3

嘗試這個：

import json, requests

r = requests.get('https://www.xkcd.com/info.0.json')
responseJSON = None
if r.status_code == 200:
    responseJSON = json.loads(r.content)

print responseJSON  # you can access values like responseJSON['img']

從這里開始，您確定JSON響應，您不妨這樣做

responseJSON = r.json()

注意：您仍然必須進行錯誤處理。