PushOverPy
Simplistic Python Pushover API implementation not requiring more than standard libraries
Requirements
Python 2.7 (have not tested on 2.6 or 3) urllib httplib json io
Optional
Redis-py (to store API token & key) xml (if you prefer XML responses, will have to write it in yourself, however)
Purpose
This was written to support the full range of features that the Pushover API has.
Beyond that, it was also a side project I decided to take on.
Usage
import pushover
# First run with Redis installed (or if Redis is not installed, everytime)
ps = pushover('api_token', 'api_key')
# Afterwards, as long as the data is still in Redis
ps = pushover()
# To store the data in Redis afterwards
ps.store_api()
# To send a message with all defaults
ps.sendmsg('some message')
# To send a message one character at a time
ps.sendmsg('some message', blocksize=1)
# High priority
ps.sendmsg('msg', priority=1)
# Low priority
ps.sendmsg('msg', priority=-1)
# Custom timestamp ('timestamp' = Epoch value)
ps.sendmsg('msg', timestamp=0)
# Custom title for push
ps.sendmsg('msg', title='Customized Title')
# Send URL with push
ps.sendmsg('msg', url='http://www.google.com', url_title='Google Search')On success, sendmsg() will return 1. Otherwise, if there's any error messages explaining why it failed, that will be returned. Lastly, if no error messages than it'll return 0.