blocks|key|415855|text|如果URL返回的是有效的JSON编码数据，则使用json+library对其进行解码：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|415856|import+urllib2
import+json

response+=+urllib2.urlopen('https://api.instagram.com/v1/tags/pizza/media/XXXXXX')
data+=+json.load(response)+++
print+data|code-block|syntax|javascript|415857|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/2/library/json.html^0|O|4|O|C|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

If the URL is returning valid JSON-encoded data, use the <a href="http://docs.python.org/2/library/json.html" rel="noreferrer"><code>json</code> library</a> to decode that:

<pre><code>import urllib2
import json

response = urllib2.urlopen('https://api.instagram.com/v1/tags/pizza/media/XXXXXX')
data = json.load(response) 
print data
</code></pre>

blocks|key|197684|text|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)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|197685|urllib，用于Python3.4|unstyled|offset|length|197686|由r.info()返回的HTTPMessage|197687|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3.4/library/urllib.html|1|https://docs.python.org/3.4/library/email.message.html#email.message.Message.get_param^0|0|0|6|0|0|C|B|1|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|V|8|@]|9|@$G|W|H|X|1|Y]]|A|$]]|$1|I|3|J|5|F|7|Z|8|@]|9|@$G|10|H|11|1|12]]|A|$]]|$1|K|3|-4|5|F|7|13|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]|S|$5|N|O|P|A|$Q|T]]]]

<pre><code>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)
</code></pre>

<a href="https://docs.python.org/3.4/library/urllib.html" rel="noreferrer" title="urllib">urllib</a>, for Python 3.4 
<a href="https://docs.python.org/3.4/library/email.message.html#email.message.Message.get_param" rel="noreferrer" title="HTTPMessage">HTTPMessage</a>, returned by r.info()

blocks|key|415943|text|"""
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")|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|415944|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>"""
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")
</code></pre>

blocks|key|197649|text|要注意验证等，但直接的解决方案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|197650|import+json
the_dict+=+json.load(response)|code-block|syntax|javascript|197651|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Be careful about the validation and etc, but the straight solution is this:

<pre><code>import json
the_dict = json.load(response)
</code></pre>

blocks|key|197689|text|resource_url+=+'http://localhost:8080/service/'
response+=+json.loads(urllib2.urlopen(resource_url).read())|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|197690|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>resource_url = 'http://localhost:8080/service/'
response = json.loads(urllib2.urlopen(resource_url).read())
</code></pre>

blocks|key|415957|text|Python+3标准库一行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|415958|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'|code-block|syntax|javascript|415959|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Python 3 standard library one-liner:

<pre class="lang-py prettyprint-override"><code>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'
</code></pre>

blocks|key|415994|text|您也可以使用requests获取json，如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|415995|import+requests

r+=+requests.get('http://yoursite.com/your-json-pfile.json')
json_response+=+r.json()|code-block|syntax|javascript|415996|entityMap^0|6|8|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

you can also get json by using <code>requests</code> as below:

<pre><code>import requests

r = requests.get('http://yoursite.com/your-json-pfile.json')
json_response = r.json()
</code></pre>

blocks|key|416025|text|这是你问题的另一个更简单的解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|416026|pd.read_json(data)|code-block|syntax|javascript|416027|其中data是以下代码的字符串输出|416028|response+=+urlopen("https://data.nasa.gov/resource/y77d-th95.json")
json_data+=+response.read().decode('utf-8',+'replace')|416029|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

This is another simpler solution to your question

<pre><code>pd.read_json(data)
</code></pre>

where data is the str output from the following code

<pre><code>response = urlopen("https://data.nasa.gov/resource/y77d-th95.json")
json_data = response.read().decode('utf-8', 'replace')
</code></pre>

blocks|key|415931|text|这里提供的示例对我来说都不起作用。它们要么用于Python2+(uurllib2)，要么用于Python3返回错误"ImportError:+No+module+named+request“。我在谷歌上搜索了错误信息，它显然需要我安装一个模块--对于这样一个简单的任务，这显然是不可接受的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|415932|这段代码对我来说很有效：|415933|import+json,urllib
data+=+urllib.urlopen("https://api.github.com/users?since=0").read()
d+=+json.loads(data)
print+(d)|code-block|syntax|javascript|415934|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

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. 

This code worked for me: 

<pre><code>import json,urllib
data = urllib.urlopen("https://api.github.com/users?since=0").read()
d = json.loads(data)
print (d)
</code></pre>

I am trying to GET a URL using Python and the response is JSON. However, when I run

<pre><code>import urllib2
response = urllib2.urlopen('https://api.instagram.com/v1/tags/pizza/media/XXXXXX')
html=response.read()
print html
</code></pre>

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.

Python urllib2: Receive JSON response from url

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在尝试使用Python获取URL，得到的响应是JSON。但是，当我运行import urllib2response = urllib2.urlopen('https://api.instagram.com/v1/tags/pizza/media/XXXXXX')html=response.read()print htmlhtml的类型是str，而我期望的是JSON。有没有办法将响应捕获为JSO

问Python urllib2:从url接收JSON响应
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