Stack.PY - A Python Module for Accessing the Stack Exchange 2.1 API

Question

Now on PyPI!

You can now find Stack.PY on PyPI, Python's package index. This means that you can install the package simply by running the following command in a terminal:

pip install stackpy

About

Stack.PY is based heavily on Stack.PHP and Stack.JS, taking the chained-method concept and applying it to Python. The end result is an extremely easy to use module, named stackpy:

from stackpy import API, Site

# Print the names of all Stack Exchange sites
for site in API.sites:
    print(site['name'])

# Grab the first question on Stack Overflow
print(Site('stackoverflow').questions[0].title)

Here are some of the other features you can expect from Stack.PY:

• Caching: currently Stack.PY ships with an SQLite database backend (used for caching currently). By default, if no cache is set, Stack.PY creates an in-memory SQLite database to cache requests for the current session.
• Full documentation: using a single command (see the README file) you can generate all of the documentation for the entire module - including an explanation for each parameter of every method.

Many more features are planned:

• A test suite.
• A series of examples (currently there is one really primitive example).
• Full support for rate-limiting and the backoff response.

Stack.PY should run perfectly fine in Python 3k using 2to3.

License

Stack.PY is released under the MIT License.

Contact

I can be reached at [email protected].

Code

The code for Stack.PY is hosted here on Launchpad. You can check out the latest code using:

bzr init
bzr pull lp:stackpy

You can view the code online here.

Stack.PY uses distutils so you can install the module by running:

python setup.py install

...or... if you are using Ubuntu, you can add my PPA and install the appropriate package:

sudo apt-get install python-stackpy (for Python 2k)
sudo apt-get install python3-stackpy (for Python 3k)

Answer 1

Example of Explicit Authentication:

Explicit authentication is relatively straightforward with Stack.PY. Assuming you have already registered your application, the instructions look something like this:

1. Visit your application's page to obtain the following information:

   • key
   • client ID
   • client secret
     
     

2. Provide Stack.PY with the above information:

   from stackpy import API
   
   # Fill in the strings with the appropriate values:
   API.key, API.client_id, API.client_secret = "", "", ""
   

3. The API.begin_explicit() call consists of the following parameters:

   • a string consisting of the required privileges separated by commas
   • the URI to redirect the user to when the authorization completes
   • an optional string value to be returned with the access token when authorization completes
     
     

   Example:

   redirect_uri = API.begin_explicit('read_inbox,no_expiry',
                                     'http://example.com/done')
   

   The return value of the function is the URL that you will need to redirect the user to. That page will allow the user to authorize your application.

4. The API.complete_explicit() call consists of the following parameters:

   • the value of the GET parameter code
   • the URI you provided to begin_explicit() above
     
     

   Example:

   access_token = API.complete_explicit(request.GET['code'],
                                        'http://example.com/done')
   

   The return value is the access token for the user.

Answer 2

Example - Downloading a question with its body

from stackpy import Site

site = Site('stackoverflow')
question = site.questions(1732348).filter('withbody')[0]

print '--- %s ---' % question.title
print question.body

Answer 3

Question: How do you request all Tags?

Retrieving a list of tags on a given Stack Exchange site with Stack.PY is relatively straightforward:

from stackpy import Site

sa = Site('stackapps')
for t in sa.tags:
    print t

However, this will only fetch the first 30 tags. The tricky part is fetching more than one page. You can fetch any particular page by sticking .page(n) on the end of the request chain (where n is the page to fetch).

If we rewrite the loop a bit, we end up with:

from itertools import count
from stackpy import Site

sa = Site('stackapps')
for p in count(1):
    tags = sa.tags.page(p)
    for t in tags:
        print t
    if not tags['has_more']:
        break

The example above uses a generator that will yield an infinite list of consecutive page numbers to fetch all of the pages. When the has_more property is set to false in the JSON returned by the API, the loop will terminate.

Answer 4

Do you know, offhand, how complete this library is? Are there parts of the StackExchange API v2.1 that you know are not covered yet?

Answer 5

After I installed stackpy, how can I get the docs? I installed doxygen but cannot get Stackpy documentation. Can somebody help me? In fact, I want to know how to search StackOverflow for related questions in my own QA site using your StackExchange API app.

Many thanks.

Answer 6

My actual requirement is to find the answers to question with specified tags. That is given a tag I want to get answers to questions with those tag. I am hoping when I get the tag I will get the time of answering as well. How can I do that. More over I am getting the below error. Can anyone help me with this.

    Traceback (most recent call last):
  File "C:\Users\ANI\workspace\StackOverFlow\src\Pydev.py", line 28, in <module>
    question = sa.answers(ans)[0]
  File "C:\Users\ANI\Anaconda-1.9.2\lib\site-packages\stackpy\request.py", line 135, in __getitem__
    return self._fetch()['items'][index] if type(index) == int else self._fetch()[index]
  File "C:\Users\ANI\Anaconda-1.9.2\lib\site-packages\stackpy\request.py", line 157, in _fetch
    self._data = self._url.fetch()
  File "C:\Users\ANI\Anaconda-1.9.2\lib\site-packages\stackpy\url.py", line 87, in fetch
    raise APIError(data['error_id'], data['error_message'])
stackpy.url.APIError: API error 502: too many requests from this IP, more requests available in 84392 seconds.