I'm making a request in python to a web service which returns AMF. I don't know if it's AMF0 or AMF3 yet.
r = requests.post(url, data=data)
>>> r.text
u'\x00\x03...'
(Full data here)
How can I take r.text and convert it to a python object or similar? I found amfast but it's Decoder class returns a 3.131513074181806e-294 assuming AMF0 and None for AMF3. (Both incorrect)
from amfast.decoder import Decoder
decoder = Decoder(amf3=False)
obj = decoder.decode(StringIO.StringIO(r.text))
I see two problems with your code, the first is using r.text to return binary data. Use r.content instead. The second problem is using decoder.decode method, which decodes one object and not a packet. Use decoder.decode_packet instead.
from amfast.decoder import Decoder
decoder = Decoder(amf3=True)
obj = decoder.decode_packet(r.content)
Using Pyamf works as well, using r.content.
have u tried PyAMF.
from pyamf import remoting
remoting.decode(data)
Related
I'm trying to read a website's content but I get an empty bytes object, b''.
import urllib3
from urllib3 import PoolManager
urllib3.disable_warnings()
https = PoolManager()
r = https.request('GET', 'https://minemen.club/leaderboards/practice/')
print(r.status)
print(r.read())
When I open the URL in a web browser I see the website, and r.status is 200 (success).
Why does r.read() not return the content?
What makes you think it is wrong? Try the following, you'll have much more output:
print(r.data)
Check HTTPResponse to see how to use the r object you got.
This is how urllib3.response.HTTPResponse.read is supposed to work.
It is explained for example here by one of the contributors to urllib3:
This is about documentation. You cannot use read() by default, because
by default all the content is consumed into data. If you want read()
to work, you need to set preload_content=True on the call to urlopen.
Want to give that a try?
So you can simply use r.data.
I'm having problems getting data from an HTTP response. The format unfortunately comes back with '\n' attached to all the key/value pairs. JSON says it must be a str and not "bytes".
I have tried a number of fixes so my list of includes might look weird/redundant. Any suggestions would be appreciated.
#!/usr/bin/env python3
import urllib.request
from urllib.request import urlopen
import json
import requests
url = "http://finance.google.com/finance/info?client=ig&q=NASDAQ,AAPL"
response = urlopen(url)
content = response.read()
print(content)
data = json.loads(content)
info = data[0]
print(info)
#got this far - planning to extract "id:" "22144"
When it comes to making requests in Python, I personally like to use the requests library. I find it easier to use.
import json
import requests
r = requests.get('http://finance.google.com/finance/info?client=ig&q=NASDAQ,AAPL')
json_obj = json.loads(r.text[4:])
print(json_obj[0].get('id'))
The above solution prints: 22144
The response data had a couple unnecessary characters at the head, which is why I am only loading the relevant (json) portion of the response: r.text[4:]. This is the reason why you couldn't load it as json initially.
Bytes object has method decode() which converts bytes to string. Checking the response in the browser, seems there are some extra characters at the beginning of the string that needs to be removed (a line feed character, followed by two slashes: '\n//'). To skip the first three characters from the string returned by the decode() method we add [3:] after the method call.
data = json.loads(content.decode()[3:])
print(data[0]['id'])
The output is exactly what you expect:
22144
JSON says it must be a str and not "bytes".
Your content is "bytes", and you can do this as below.
data = json.loads(content.decode())
I was following a tutorial about how to use JSON objects (link: https://www.youtube.com/watch?v=Y5dU2aGHTZg). When they ran the code, they got no errors, but I did. Is it something to do with different Python versions or something?
from urllib.request import urlopen
import json
def printResults(data):
theJSON = json.loads(data)
print (theJSON)
def main():
urlData ="http://earthquake.usgs.gov/earthquakes/feed/v1.0/summary/2.5_day.geojson"
webUrl = urlopen(urlData)
print(webUrl.getcode())
if (webUrl.getcode()==200):
data = webUrl.read()
printResults(data)
else:
print ("You failed")
main()
The HTTPResponse object returned from urlopen reads bytes data (raw binary data), not str data (textual data), while the json module works with str. You need to know (or inspect the headers to determine) the encoding used for the data received, and decode it appropriately before using json.loads.
Assuming it's UTF-8 (most websites are), you can just change:
data = webUrl.read()
to:
data = webUrl.read().decode('utf-8')
and it should fix your problem.
I think they were using a different version of the urllib
Try with urllib3 and do the import like this:
from urllib import urlopen
Hope this is the fix to your problem
I am trying to GET a URL using Python and the response is JSON. However, when I run
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.
If the URL is returning valid JSON-encoded data, use the json library to decode that:
import urllib2
import json
response = urllib2.urlopen('https://api.instagram.com/v1/tags/pizza/media/XXXXXX')
data = json.load(response)
print data
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()
"""
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")
Be careful about the validation and etc, but the straight solution is this:
import json
the_dict = json.load(response)
resource_url = 'http://localhost:8080/service/'
response = json.loads(urllib2.urlopen(resource_url).read())
Python 3 standard library one-liner:
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'
you can also get json by using requests as below:
import requests
r = requests.get('http://yoursite.com/your-json-pfile.json')
json_response = r.json()
Though I guess it has already answered I would like to add my little bit in this
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
This is another simpler solution to your question
pd.read_json(data)
where data is the str output from the following code
response = urlopen("https://data.nasa.gov/resource/y77d-th95.json")
json_data = response.read().decode('utf-8', 'replace')
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:
import json,urllib
data = urllib.urlopen("https://api.github.com/users?since=0").read()
d = json.loads(data)
print (d)
Looking for a python script that would simply connect to a web page (maybe some querystring parameters).
I am going to run this script as a batch job in unix.
urllib2 will do what you want and it's pretty simple to use.
import urllib
import urllib2
params = {'param1': 'value1'}
req = urllib2.Request("http://someurl", urllib.urlencode(params))
res = urllib2.urlopen(req)
data = res.read()
It's also nice because it's easy to modify the above code to do all sorts of other things like POST requests, Basic Authentication, etc.
Try this:
aResp = urllib2.urlopen("http://google.com/");
print aResp.read();
If you need your script to actually function as a user of the site (clicking links, etc.) then you're probably looking for the python mechanize library.
Python Mechanize
A simple wget called from a shell script might suffice.
in python 2.7:
import urllib2
params = "key=val&key2=val2" #make sure that it's in GET request format
url = "http://www.example.com"
html = urllib2.urlopen(url+"?"+params).read()
print html
more info at https://docs.python.org/2.7/library/urllib2.html
in python 3.6:
from urllib.request import urlopen
params = "key=val&key2=val2" #make sure that it's in GET request format
url = "http://www.example.com"
html = urlopen(url+"?"+params).read()
print(html)
more info at https://docs.python.org/3.6/library/urllib.request.html
to encode params into GET format:
def myEncode(dictionary):
result = ""
for k in dictionary: #k is the key
result += k+"="+dictionary[k]+"&"
return result[:-1] #all but that last `&`
I'm pretty sure this should work in either python2 or python3...
What are you trying to do? If you're just trying to fetch a web page, cURL is a pre-existing (and very common) tool that does exactly that.
Basic usage is very simple:
curl www.example.com
You might want to simply use httplib from the standard library.
myConnection = httplib.HTTPConnection('http://www.example.com')
you can find the official reference here: http://docs.python.org/library/httplib.html