Develop Reference

Python request gives 415 error while post data unless data = {'key':'value'}

I am doing a straight forward request as follows.
import requests
def user_transactions():
url = 'https://webapi.coinfloor.co.uk/v2/bist/XBT/GBP/user_transactions/'
data = {'key':'value'}
r = requests.post(url, data=data, auth=("some_username", "some_password") )
print(r.status_code)
print(r.text)
return
Even though data= is optional in the documents.
https://www.w3schools.com/python/ref_requests_post.asp
If i comment out the data variable then the routine returns a
status_code=415 error.
If i include in the data variable then the routine returns a status_code=200 success.
I have tried to look this up, for example here:
Python request gives 415 error while post data , but with no answer.
The question is: Why is it the case that [1] fails but [2] works ?

Yes, data is optional on the python side. The requests library will happily send a empty request to the server, as you can see. If the argument was not optional, the program would crash before sending a request so there would be no status code.
However, the server needs to be able to process the request. If it does not like what you sent for whatever reason, it might send back a 4xx status code, or otherwise not do what you expect.
In this case, it throws an error that the data is in invalid format. How can a empty request be in invalid format? Because the format is specified in a header. If you supply a data argumet requests will send data in urlencoded format, and specify in the header what format the data is in. If the data is empty, the request will be empty but the header will still be there. This site apparently requires the header to specify a data format it knows.
You can solve this in two ways, giving an empty object:
r = requests.post(url, data={}, auth=("some_username", "some_password") )
Or by explicitly specifying the header:
r = requests.post(url, auth=(...), headers={'Content-Type': 'application/x-www-form-urlencoded'})
Side note: You should not be using W3Schools as a source. It is frequently inaccurate and often recommends bad practices.

I think you are mistaking the documentation of the requests.post function signature with API documentation. It is saying that data is a keyword argument, not that the API optionally takes data.
It depends on the API endpoint you are trying to use. That endpoint must require data to be sent with the request. If you look at the documentation for the API you are using, it will mention what needs to be sent for a valid request.

How do I make this website recognise my arrays as part of a valid url query?

EDIT:
In a similar vein, when I now try to log into their account with a post request, what is returned is none of the errors they suggest on their site, but is in fact a "JSON exception". Is there any way to debug this, or is an error code 500 completely impossible to deal with?
I'm well aware this question has been asked before. Sadly, when trying the proposed answers, none worked. I have an extremely simple Python project with urllib, and I've never done web programming in Python before, nor am I even a regular Python user. My friend needs to get access to content from this site, but their user-friendly front-end is down and I learned that they have a public API to access their content. Not knowing what I'm doing, but glad to try to help and interested in the challenge, I have very slowly set out.
Note that it is necessary for me to only use standard Python libraries, so that any finished project could easily be emailed to their computer and just work.
The following works completely fine minus the "originalLanguage" query, but when using it, which the API has documented as an array value, no matter whether I comma-separate things, or write "originalLanguage[0]" or "originalLanguage0" or anything that I've seen online, this creates the error message from the server: "Array value expected but string detected" or something along those lines.
Is there any way for me to get this working? Because it clearly can work, otherwise the API wouldn't document it. Many thanks.
In case it helps, when using "[]" or "<>" or "{}" or any delimeter I could think of, my IDE didn't recognise it as part of the URL.
import urllib.request as request
import urllib.parse as parse
def make_query(url, params):
url += "?"
for i in range(len(params)):
url += list(params)[i]
url += '='
url += list(params.values())[i]
if i < len(params) - 1:
url += '&'
return url
base = "https://api.mangadex.org/manga"
params = {
"limit": "50",
"originalLanguage": "en"
}
url = make_query(base, params)
req = request.Request(url)
response = request.urlopen(req)

Why do I get this error - "missing token"? (playing with an API using Python)

So, I am playing around with Etilbudsavis' API (Danish directory containing offers from retail stores). My goal is to retrieve data based on a search query. the API acutally allows this, out of the box. However, when I try to do this, I end up with an error saying that my token is missing. Anyways, here is my code:
from urllib2 import urlopen
from json import load
import requests
body = {'api_key': 'secret_api_key'}
response = requests.post('https://api.etilbudsavis.dk/v2/sessions', data=body)
print response.text
new_body = {'_token:': 'token_obtained_from_POST_method', 'query:': 'coca-cola'}
new_response = requests.get('https://api.etilbudsavis.dk/v2/offers/search', data=new_body)
print new_response.text
Full error:
{"code":1107,"id":"00ilpgq7etum2ksrh4nr6y1jlu5ng8cj","message":"missing token","
details":"Missing token\nNo token found in request to an endpoint that requires
a valid token.","previous":null,"#note.1":"Hey! It looks like you found an error
. If you have any questions about this error, feel free to contact support with
the above error id."}

Since this is a GET request, you should use the params argument to pass the data in the URL.
new_response = requests.get('https://api.etilbudsavis.dk/v2/offers/search', params=new_body)
See the requests docs.

I managed to solve the problem with the help of Daniel Roseman who reminded me of the fact that playing with an API in the Python Shell is different from interacting with the API in the browser. The docs clearly stated that you'd have to sign the API token in order for it to work. I missed that tiny detail ... Never the less, Daniel helped me figure everything out. Thanks again, Dan.

Urllib's urlopen breaking on some sites (e.g. StackApps api): returns garbage results

I'm using urllib2's urlopen function to try and get a JSON result from the StackOverflow api.
The code I'm using:
>>> import urllib2
>>> conn = urllib2.urlopen("http://api.stackoverflow.com/0.8/users/")
>>> conn.readline()
The result I'm getting:
'\x1f\x8b\x08\x00\x00\x00\x00\x00\x04\x00\xed\xbd\x07`\x1cI\x96%&/m\xca{\x7fJ\...
I'm fairly new to urllib, but this doesn't seem like the result I should be getting. I've tried it in other places and I get what I expect (the same as visiting the address with a browser gives me: a JSON object).
Using urlopen on other sites (e.g. "http://google.com") works fine, and gives me actual html. I've also tried using urllib and it gives the same result.
I'm pretty stuck, not even knowing where to look to solve this problem. Any ideas?

That almost looks like something you would be feeding to pickle. Maybe something in the User-Agent string or Accepts header that urllib2 is sending is causing StackOverflow to send something other than JSON.
One telltale is to look at conn.headers.headers to see what the Content-Type header says.
And this question, Odd String Format Result from API Call, may have your answer. Basically, you might have to run your result through a gzip decompressor.
Double checking with this code:
>>> req = urllib2.Request("http://api.stackoverflow.com/0.8/users/",
headers={'Accept-Encoding': 'gzip, identity'})
>>> conn = urllib2.urlopen(req)
>>> val = conn.read()
>>> conn.close()
>>> val[0:25]
'\x1f\x8b\x08\x00\x00\x00\x00\x00\x04\x00\xed\xbd\x07`\x1cI\x96%&/m\xca{\x7fJ'
Yes, you are definitely getting gzip encoded data back.
Since you seem to be getting different results on different machines with the same version of Python, and in general it looks like the urllib2 API would require you do something special to request gzip encoded data, my guess is that you have a transparent proxy in there someplace.
I saw a presentation by the EFF at CodeCon in 2009. They were doing end-to-end connectivity testing to discover dirty ISP tricks of various kinds. One of the things they discovered while doing this testing is that a surprising number of consumer level NAT routers add random HTTP headers or do transparent proxying. You might have some piece of equipment on your network that's adding or modifying the Accept-Encoding header in order to make your connection seem faster.

Python urllib2 Response header

I'm trying to extract the response header of a URL request. When I use firebug to analyze the response output of a URL request, it returns:
Content-Type text/html
However when I use the python code:
urllib2.urlopen(URL).info()
the resulting output returns:
Content-Type: video/x-flv
I am new to python, and to web programming in general; any helpful insight is much appreciated. Also, if more info is needed please let me know.
Thanks in advance for reading this post

Try to request as Firefox does. You can see the request headers in Firebug, so add them to your request object:
import urllib2
request = urllib2.Request('http://your.tld/...')
request.add_header('User-Agent', 'some fake agent string')
request.add_header('Referer', 'fake referrer')
...
response = urllib2.urlopen(request)
# check content type:
print response.info().getheader('Content-Type')
There's also HTTPCookieProcessor which can make it better, but I don't think you'll need it in most cases. Have a look at python's documentation:
http://docs.python.org/library/urllib2.html

Content-Type text/html
Really, like that, without the colon?
If so, that might explain it: it's an invalid header, so it gets ignored, so urllib guesses the content-type instead, by looking at the filename. If the URL happens to have ‘.flv’ at the end, it'll guess the type should be video/x-flv.

This peculiar discrepancy might be explained by different headers (maybe ones of the accept kind) being sent by the two requests -- can you check that...? Or, if Javascript is running in Firefox (which I assume you're using when you're running firebug?) -- since it's definitely NOT running in the Python case -- "all bets are off", as they say;-).

Keep in mind that a web server can return different results for the same URL based on differences in the request. For example, content-type negotiation: the requestor can specify a list of content-types it will accept, and the server can return different results to try to accomodate different needs.
Also, you may be getting an error page for one of your requests, for example, because it is malformed, or you don't have cookies set that authenticate you properly, etc. Look at the response itself to see what you are getting.

according to http://docs.python.org/library/urllib2.html there is only get_header() method and nothing about getheader .
Asking because Your code works fine for
response.info().getheader('Set cookie')
but once i execute
response.info().get_header('Set cookie')
i get:
Traceback (most recent call last):
File "baza.py", line 11, in <module>
cookie = response.info().get_header('Set-Cookie')
AttributeError: HTTPMessage instance has no attribute 'get_header'
edit:
Moreover
response.headers.get('Set-Cookie') works fine as well, not mentioned in urlib2 doc....

for getting raw data for the headers in python2, a little bit of a hack but it works.
"".join(urllib2.urlopen("http://google.com/").info().__dict__["headers"])
basically "".join(list) will the list of headers, which all include "\n" at the end.
__dict__ is a built in python variable for all dicts, basically you can select a list out of a 2d array with it.
and ofcourse ["headers"] is selecting the list value from the .info() response value dict
hope this helped you learn a few ez python tricks :)