在python中存储这个JSON的正确方法是什么？

Question

我试图存储JSON对象从网站https://www.xkcd.com/info.0.json

我已经试过

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)回报

{ 
    '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." 
} 

要将单引号转换成response_content返回，我尝试过

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

但问题与线出现在那里alt是

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

如果里面有任何值的单引号，这种方法失败。

错误日志：

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) 

加载的JSON在脚本中任何其他的方法呢？

编辑：

我要像做

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

但这返回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数据结构。你在这里做的很多，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() 

就是这样。

您正在将Python数据结构转换为字符串，然后尝试再次将该字符串解释为JSON。这可能看起来像它的工作，因为Python容器的str()转换使用Python语法来产生结果。但是Python不是JSON，不管怎样，你试图把它变成JSON也是不太好的。并且根本不需要。

您可以直接使用json_data，它是一个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

试试这个：

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'] 

因为在这里，你肯定有JSON响应的，你不妨做

responseJSON = r.json() 

注意：您仍必须做错误处理。

Answer 3

response.json()已经返回一个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 

无需转换和从字符串。