plamoni / SiriProxy

A (tampering) proxy server for Apple's Siri
GNU General Public License v3.0
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Call my custon python program in Siriproxy #518

Closed HaTe87 closed 11 years ago

HaTe87 commented 11 years ago

Hi, i have installed Rupy to call a python program in ruby on my SiriProxy , but when i try to callback my program ( /root/p.py) i recived an error and i'm not able to understand it. i have put : require 'rupy' and listen_for /movimento/i do say "Vado!" #say something to the user! request_completed # RubyPython.start # start the Python interpreter cPickle = RubyPython.import("cPickle") puts cPickle.dumps("Testing RubyPython.").rubify RubyPython.stop # stop the Python interpreter end in siriproxy-example.rb but if i dont' put # before require the proxy dosen't work if i put the # i recived [Info - iPhone] Received Object: FinishSpeech [Info - Guzzoni] Received Object: SpeechRecognized [Info - Plugin Manager] Processing 'Movimento' [Info - Plugin Manager] Processing plugin Example [Info - Plugin Manager] Matches (?i-mx:movimento) [Info - Plugin Manager] Applicable states: [Info - Plugin Manager] Current state: [Info - Plugin Manager] Matches, executing block [Info - Plugin Manager] Say: Vado! [Info - Plugin Manager] Sending Request Completed [Info - Plugin Manager] Plugin Crashed: uninitialized constant SiriProxy::Plugin::Example::RubyPython can anyone help me?

elvisimprsntr commented 11 years ago

This issues tracker is for problems with SiriProxy itself, not an open forum for how to program in Ruby. I suggest you seek a Ruby specific forum for your question, ie stackoverflow. Regardless, if it were me I would change cPickle variable to all lower case, ie. cpickle