I'm trying to call the NYT events api with urllib2 but I'm receiving a 596 error. If I construct the url myself, there is no problem, but if I call urlopen with the data instead, I receive the 596 error. What's going on? The 596 error seems to be undocumented, so it doesn't help.
>>> data = urllib.urlencode({'api-key': os.environ['NYT_EVENT_LISTING_API_KEY']})
>>> resp = urllib2.urlopen('?'.join([url,data]))
>>> resp = urllib2.urlopen(url, data)
Traceback (most recent call last):
File "<input>", line 1, in <module>
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 126, in urlopen
return _opener.open(url, data, timeout)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 400, in open
response = meth(req, response)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 513, in http_response
'http', request, response, code, msg, hdrs)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 438, in error
return self._call_chain(*args)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 372, in _call_chain
result = func(*args)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 521, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
HTTPError: HTTP Error 596:
#Thomas is right, you are using GET in your first request which will construct the URL to something like this:
nytimes.com/api/?MY_API_KEY
However, your second call to urllib2.urlopen sends the data as a POST request to this URL
nytimes.com/api/
instead, which gives you 596 service not found error.
Now, urllib2 is notorious for its non-intuitive API and documentation, you may consider using Requests instead:
import requests
api_key = {'api-key': os.environ['NYT_EVENT_LISTING_API_KEY']}
resp = requests.get(url, params=api_key)
print resp.text
print resp.json
This way, GET requests and POST requests are a lot easier to distinguish, url and parameters are separated as well.
Your first request is a GET request - the second is a POST request. See the docs on this - when the parameter data is provided, urlopen performs a POST request.
Related
Problem
I'm using urllib.request.urlopen on the Wall Street Journal and it gives me a 404.
Details
Other sites work fine. Same error if I use https://. I did this example in REPL but the same error happens in my calls from my Django server:
>>> from urllib.request import urlopen
>>> urlopen('http://www.wsj.com')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/urllib/request.py", line 222, in urlopen
return opener.open(url, data, timeout)
File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/urllib/request.py", line 531, in open
response = meth(req, response)
File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/urllib/request.py", line 641, in http_response
'http', request, response, code, msg, hdrs)
File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/urllib/request.py", line 569, in error
return self._call_chain(*args)
File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/urllib/request.py", line 503, in _call_chain
result = func(*args)
File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/urllib/request.py", line 649, in http_error_default
raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 404: Not Found
This is how it should work:
>>> urlopen('http://www.cbc.ca')
<http.client.HTTPResponse object at 0x10b0f8c88>
I'm not sure how to debug this. Anyone know what's going on, and how I can fix it?
first import Request like this:
from urllib.request import **Request**, urlopen
and then pass your url and header to Request like below:
url = 'https://www.wsj.com/'
response_obj = urlopen(Request(url, headers={'User-Agent': 'Mozilla/5.0'}))
print(response_obj)
I tested it now its working
Mod security blocks me from accessing a URL.
I have tried using other methods from this website, but none seem to work as I use an import called PyCurrency-Calculator
import PyCurrency_Converter
import urllib.request
from bs4 import BeautifulSoup
PyCurrency_Converter.convert(1, 'USD', 'A$')
The error is:
Traceback (most recent call last):
File "E:/Downloads/list of currencies.py", line 7, in <module>
PyCurrency_Converter.convert(1, 'USD', 'A$')
File "E:\Python34\lib\site-packages\PyCurrency_Converter\PyCurrency.py",
line 48, in convert
return PyCurrency.convert(amount, _from, _to)
File "E:\Python34\lib\site-packages\PyCurrency_Converter\PyCurrency.py",
line 31, in convert
response = urllib2.urlopen(url)
File "E:\Python34\lib\urllib\request.py", line 161, in urlopen
return opener.open(url, data, timeout)
File "E:\Python34\lib\urllib\request.py", line 469, in open
response = meth(req, response)
File "E:\Python34\lib\urllib\request.py", line 579, in http_response
'http', request, response, code, msg, hdrs)
File "E:\Python34\lib\urllib\request.py", line 507, in error
return self._call_chain(*args)
File "E:\Python34\lib\urllib\request.py", line 441, in _call_chain
result = func(*args)
File "E:\Python34\lib\urllib\request.py", line 587, in http_error_default
raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 403: Forbidden
PyCurrency_Converter makes a call google finance under the hood (code). This service is no longer available and so the library needs to be modified to use a different service.
What you can do is find an alternative web based api and call that (I have no recommendation). As you can see the code for PyCurrency_Converter is quite short, so it will be simple to take and adapt it to another currency conversion api or to use requests or urllib to just do the requests yourself.
I'm trying to use the yts api, and I wrote the following code.
import urllib2
import json
url = 'https://yts.ag/api/v2/list_movies.json?quality=3D'
json_obj = urllib2.urlopen(url)
data = json.load(json_obj)
print data
But I ran into the following error:
File "<stdin>", line 1, in <module>
File "C:\Python27\lib\urllib2.py", line 154, in urlopen
return opener.open(url, data, timeout)
File "C:\Python27\lib\urllib2.py", line 435, in open
response = meth(req, response)
File "C:\Python27\lib\urllib2.py", line 548, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Python27\lib\urllib2.py", line 473, in error
return self._call_chain(*args)
File "C:\Python27\lib\urllib2.py", line 407, in _call_chain
result = func(*args)
File "C:\Python27\lib\urllib2.py", line 556, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 403: Forbidden
Help me how to fix this
I'm not completely sure what's going on with your urllib2 request, as it works fine with the other urls I tested, with and without JSON responses.
I suspect it's an encoding thing for the quality=3D query, but my brief attempt at handling that also failed.
Simply putting the request in a browser works, as does the following 'requests' code.
In a nutshell, I'm not truly answering your question regarding fixing the urllib2 usage, but definitely fixing the problem of getting the data.
I'd suggest working with the 'requests' module in this case. Try
import requests
import json
url = 'https://yts.ag/api/v2/list_movies.json?quality=3D'
response = requests.get( url = url )
data = response.json()
print data
Note that the requests response.json() handles the json.load() for you.
It will save you many a headache over urllib2.
The 'requests' documentation:
http://docs.python-requests.org/en/latest/
I am really an ETL guy trying to learn Python, please help
import urllib2
urls =urllib2.urlopen("url1","url2")
i=0
while i< len(urls):
htmlfile = urllib2.urlopen(urls[i])
htmltext = htmlfile.read()
print htmltext
i+=1
I am getting errors as
Traceback (most recent call last):
File ".\test.py", line 2, in
urls =urllib2.urlopen("url1","url2")
File "c:\python27\Lib\urllib2.py", line 154, in urlopen
return opener.open(url, data, timeout)
File "c:\python27\Lib\urllib2.py", line 437, in open
response = meth(req, response)
File "c:\python27\Lib\urllib2.py", line 550, in http_response
'http', request, response, code, msg, hdrs)
File "c:\python27\Lib\urllib2.py", line 475, in error
return self._call_chain(*args)
File "c:\python27\Lib\urllib2.py", line 409, in _call_chain
result = func(*args)
File "c:\python27\Lib\urllib2.py", line 558, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 405: Method Not Allowed
Your error is coming from line 2:
urls =urllib2.urlopen("url1","url2")
Whatever url you're trying to access is returning a http error code
HTTP Error 405: Method Not Allowed
Looking at the urllib2 docs, you should only be using 1 url as an argument
https://docs.python.org/2/library/urllib2.html
Open the URL url, which can be either a string or a Request object.
data may be a string specifying additional data to send to the server, or None if no such data is needed. Currently HTTP requests are the only ones that use data; the HTTP request will be a POST instead of a GET when the data parameter is provided.
The 2nd argument you're putting in may be turning the request into a POST, which would explain the Method Not Allowed code.
Hello fellow Programmers,
today I wanted to get some JSON Data from this website using Python 3.3: http://ladv.de/api/-apikey-redacted-/ausDetail?id=884&wettbewerbe=true&all=true
The official API tells me that calling this URL returns some JSON Data. But if I use the following code to get it (which I found on stackoverflow, too), it throws an error:
import urllib.request
import json
request = 'http://ladv.de/api/mmetzger/ausDetail?id=884&wettbewerbe=true&all=true'
response = urllib.request.urlopen(request)
obj = json.load(response)
str_response = response.readall().decode('utf-8')
obj = json.loads(str_response)
print(obj)
prints out
Traceback (most recent call last):
File "D:/ladvclient/testscrape.py", line 5, in <module>
response = urllib.request.urlopen(request)
File "C:\Python33\lib\urllib\request.py", line 156, in urlopen
return opener.open(url, data, timeout)
File "C:\Python33\lib\urllib\request.py", line 475, in open
response = meth(req, response)
File "C:\Python33\lib\urllib\request.py", line 587, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Python33\lib\urllib\request.py", line 513, in error
return self._call_chain(*args)
File "C:\Python33\lib\urllib\request.py", line 447, in _call_chain
result = func(*args)
File "C:\Python33\lib\urllib\request.py", line 595, in http_error_default
raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 404: Not Found
Where is the bug, and what is the correct code?
Thanks in advance,
forumfresser
The site you're trying to fetch is not available, as seen here:
http://ladv.de/api/-apikey-redacted-/ausDetail?id=884&wettbewerbe=true&all=true
You could also just read the error message by yourself:
urllib.error.HTTPError: HTTP Error 404: Not Found