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add lock to ensure ipython execute will not be stuck forever when complete is call

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This commit is contained in:
marc hurabielle 2019-03-20 18:10:11 +09:00
parent 333a34a92f
commit 409b75f0fb
1 changed files with 9 additions and 3 deletions
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12
python/src/main/resources/grpc/python/ipython_server.py
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@ -38,6 +38,10 @@ class IPython(ipython_pb2_grpc.IPythonServicer):
def __init__(self, server):
self._status = ipython_pb2.STARTING
self._server = server
# issue with execute_interactive and auto completion: https://github.com/jupyter/jupyter_client/issues/429
# in all case because ipython does not support run and auto completion at the same time: https://github.com/jupyter/notebook/issues/3763
# For now we will lock to ensure that there is no concurrent bug that can "hang" the kernel
self._lock = threading.Lock()
def start(self):
print("starting...")
@ -82,10 +86,11 @@ class IPython(ipython_pb2_grpc.IPythonServicer):
payload_reply = []
def execute_worker():
reply = self._kc.execute_interactive(request.code,
with self._lock:
reply = self._kc.execute_interactive(request.code,
output_hook=_output_hook,
timeout=None)
payload_reply.append(reply)
payload_reply.append(reply)
t = threading.Thread(name="ConsumerThread", target=execute_worker)
t.start()
@ -169,7 +174,8 @@ class IPython(ipython_pb2_grpc.IPythonServicer):
return ipython_pb2.CancelResponse()
def complete(self, request, context):
reply = self._kc.complete(request.code, request.cursor, reply=True, timeout=None)
with self._lock:
reply = self._kc.complete(request.code, request.cursor, reply=True, timeout=None)
return ipython_pb2.CompletionResponse(matches=reply['content']['matches'])
def status(self, request, context):