Python urllib2：从 url 接收 JSON 响应
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Python urllib2: Receive JSON response from url
提问by Deepak B
I am trying to GET a URL using Python and the response is JSON. However, when I run
我正在尝试使用 Python 获取 URL，响应是 JSON。然而，当我跑
import urllib2
response = urllib2.urlopen('https://api.instagram.com/v1/tags/pizza/media/XXXXXX')
html=response.read()
print html
The html is of type str and I am expecting a JSON. Is there any way I can capture the response as JSON or a python dictionary instead of a str.
html 是 str 类型，我期待一个 JSON。有什么方法可以将响应捕获为 JSON 或 python 字典而不是 str.json 。
采纳答案by Martijn Pieters
If the URL is returning valid JSON-encoded data, use the jsonlibraryto decode that:
如果 URL 返回有效的 JSON 编码数据，请使用json库对其进行解码：
import urllib2
import json
response = urllib2.urlopen('https://api.instagram.com/v1/tags/pizza/media/XXXXXX')
data = json.load(response)
print data
回答by MostafaR
Be careful about the validation and etc, but the straight solution is this:
小心验证等，但直接的解决方案是：
import json
the_dict = json.load(response)
回答by SanalBathery
import json
import urllib
url = 'http://example.com/file.json'
r = urllib.request.urlopen(url)
data = json.loads(r.read().decode(r.info().get_param('charset') or 'utf-8'))
print(data)
urllib, for Python 3.4
HTTPMessage, returned by r.info()
urllib，用于 Python 3.4
HTTPMessage，由 r.info() 返回
回答by Jossef Harush
resource_url = 'http://localhost:8080/service/'
response = json.loads(urllib2.urlopen(resource_url).read())
回答by Uxbridge
None of the provided examples on here worked for me. They were either for Python 2 (uurllib2) or those for Python 3 return the error "ImportError: No module named request". I google the error message and it apparently requires me to install a the module - which is obviously unacceptable for such a simple task.
此处提供的示例均不适合我。它们要么用于 Python 2 (uurllib2)，要么用于 Python 3 返回错误“ImportError: No module named request”。我谷歌了错误消息，它显然要求我安装一个模块——这对于这样一个简单的任务显然是不可接受的。
This code worked for me:
这段代码对我有用：
import json,urllib
data = urllib.urlopen("https://api.github.com/users?since=0").read()
d = json.loads(data)
print (d)
回答by Nitigya Sharma
Though I guess it has already answered I would like to add my little bit in this
虽然我猜它已经回答了我想在这个中添加我的一点点
import json
import urllib2
class Website(object):
def __init__(self,name):
self.name = name
def dump(self):
self.data= urllib2.urlopen(self.name)
return self.data
def convJSON(self):
data= json.load(self.dump())
print data
domain = Website("https://example.com")
domain.convJSON()
Note : object passed to json.load()should support .read(), therefore urllib2.urlopen(self.name).read()would not work . Doamin passed should be provided with protocol in this case http
注意：传递给json.load() 的对象应该支持.read()，因此urllib2.urlopen(self.name).read()将不起作用。在这种情况下，Doamin 通过的协议应该提供给http
回答by raccoon
"""
Return JSON to webpage
Adding to wonderful answer by @Sanal
For Django 3.4
Adding a working url that returns a json (Source: http://www.jsontest.com/#echo)
"""
import json
import urllib
url = 'http://echo.jsontest.com/insert-key-here/insert-value-here/key/value'
respons = urllib.request.urlopen(url)
data = json.loads(respons.read().decode(respons.info().get_param('charset') or 'utf-8'))
return HttpResponse(json.dumps(data), content_type="application/json")
回答by Adam
Python 3 standard library one-liner:
Python 3 标准库单行：
load(urlopen(url))
# imports (place these above the code before running it)
from json import load
from urllib.request import urlopen
url = 'https://jsonplaceholder.typicode.com/todos/1'
回答by Haritsinh Gohil
you can also get json by using requestsas below:
您还可以使用requests以下方法获取 json ：
import requests
r = requests.get('http://yoursite.com/your-json-pfile.json')
json_response = r.json()
回答by Himanshu Aggarwal
This is another simpler solution to your question
这是您问题的另一个更简单的解决方案
pd.read_json(data)
where data is the str output from the following code
其中 data 是以下代码的 str 输出
response = urlopen("https://data.nasa.gov/resource/y77d-th95.json")
json_data = response.read().decode('utf-8', 'replace')
