Make status page return useful information

The flask app is now multithreaded. Only one request will handle merch
at a time through /vend, and the /status endpoint will return 200 if the
machine is available to vend, and 503 if it is currently vending

Author: abrandemuehl

machine_controller/app.py

@@ -51,6 +51,7 @@ def vend():
 
 
     statuses = []
+    merch.acquire()
     for i, item in enumerate(items):
         try:
             merch.vend(item[0], int(item[1]))
@@ -62,11 +63,25 @@ def vend():
             # goes wrong, we still need to let the client know instead of
             # throwing a 500
             statuses.append({'error': str(e), 'location': item})
+    merch.release()
 
     return jsonify(transaction_id=transaction_id, items=statuses)
 
-if __name__ == '__main__':
-    # Make sure flask runs in a single thread. Otherwise concurrent requests
-    # may cause problems with vending
-    app.run(debug=True, host='0.0.0.0', threaded=False)
+@app.route('/status', methods=['GET'])
+def status():
+    if request.headers.get('Authorization', '') != token_value:
+        abort(401)
+
+    ready = merch.inUse()
+
+    if(ready):
+        # 200 to indicate success
+        return ('', 200)
 
+    else:
+        return ('', 503)
+
+
+
+if __name__ == '__main__':
+    app.run(debug=True, host='0.0.0.0', threaded=True)

machine_controller/vend.py

@@ -34,6 +34,7 @@
 # THE SOFTWARE.
 import RPi.GPIO as GPIO
 import time
+from threading import Condition, Lock
 
 
 class Merch:
@@ -56,9 +57,22 @@ def __init__(self, debug=False):
         self.__low()
         self.__commit()
 
+        self.lock = Lock()
+
     def __del__(self):
         self.__cleanup()
 
+    def acquire(self):
+        self.lock.acquire()
+
+    def release(self):
+        self.lock.release()
+
+    def inUse(self):
+        # Trylock
+        return not self.lock.acquire(False)
+
+
     def __cleanup(self):
         ''' Clean up all of the GPIO pins '''
         GPIO.cleanup()