pyGithub maximum use of API call rate

Question

I am attempting to use the pyGithub library for accessing v3 API of github. Although this library is simple to use. I found the documentation to be very vague.

Below I am successfully getting the contents of a file with the file path and its sha. My end goal is to reduce my API calls from 3 to just 1, since I want to make use of the full 5000 API calls within the hour.

from github import Github
gh = Github(access_token) # I supply an access token here.
user = gh.get_user(owner_name) # This takes 1 API call
repo = user.get_repo(repo_name) # This takes 1 API call


file = repo.get_file_contents(filename, ref=sha) # This takes 1 API call

Does anyone know how I can pass the repo and owner name to get_file_contents() or a similar function I can use to achieve this?

Any help is appreciated.

Answer 1

You can reduce it from 3 API calls to just 2 by using get_repo() with the format 'owner_name/repo_name'

from github import Github
gh = Github(access_token) # I supply an access token here.
repo = gh.get_repo(owner_name+'/'+repo_name) # This takes 1 API call

file = repo.get_file_contents(filename, ref=sha) # This takes 1 API call

Just mentioning this here for future reference. In actuality I ended up using the requests library and forming my own api calls.

Like This:

import requests
# Import python's base64 decoder
from base64 import b64decode as decode

def GET(owner_repo,filename,sha,access_token):
    # Supply Headers
    headers = {'content-type': 'application/json', 'Authorization':'token '+access_token}
    # This function is stable so long as the github API structure does not change. 
    # Also I am using the previously mentioned combo of owner/repo.
    url = 'https://api.github.com/repos/%s/contents/%s?ref=%s' % (owner_repo, filename, sha)
    # Let's stay within the API rate limits
    url_rate_limit = 'https://api.github.com/rate_limit'
    r = requests.get(url_rate_limit, headers=headers)
    rate_limit = int(r.json()['resources']['core']['remaining'])
    if(rate_limit > 0):
        # The actual request
        r = requests.get(url, headers=headers)
        # Note: you will need to handle the any error codes yourself. 
        # I am just safe checking for '200' OK
        if(r.status_code == 200):
            content = r.json()['content']
            # Decode base64 content
            decoded_content = decode(content)
            return decoded_content

I license the above code under the MIT license.