问题描述
我有以下 lambda 代码来尝试从 AWS 上的同一 VPC 访问 Neptune。从https://docs.aws.amazon.com/neptune/latest/userguide/access-graph-gremlin-python.html复制最多。
但我收到 403 Forbidden 错误。
源代码:
from gremlin_python import statics
from gremlin_python.structure.graph import Graph
from gremlin_python.process.graph_traversal import __
from gremlin_python.process.strategies import *
from gremlin_python.process.traversal import T
from gremlin_python.driver.driver_remote_connection import DriverRemoteConnection
wss = 'wss://{}:{}/gremlin'.format(host,port)
remoteConn = DriverRemoteConnection(wss,'g')
print("wss--> {}".format(wss))
# grapch
graph = Graph()
g = graph.traversal().withRemote(remoteConn)
# get
info = g.V().hasLabel('my').outE().inV().path().toList()
print(info)
return info
错误回溯:
{
"errorMessage": "HTTP 403: Forbidden","errorType": "HTTPClientError","stackTrace": [
" File \"/var/task/gremlin_python/process/traversal.py\",line 58,in toList\n return list(iter(self))\n"," File \"/var/task/gremlin_python/process/traversal.py\",line 48,in __next__\n self.traversal_strategies.apply_strategies(self)\n",line 573,in apply_strategies\n traversal_strategy.apply(traversal)\n"," File \"/var/task/gremlin_python/driver/remote_connection.py\",line 149,in apply\n remote_traversal = self.remote_connection.submit(traversal.bytecode)\n"," File \"/var/task/gremlin_python/driver/driver_remote_connection.py\",line 56,in submit\n result_set = self._client.submit(bytecode,request_options=self._extract_request_options(bytecode))\n"," File \"/var/task/gremlin_python/driver/client.py\",line 127,in submit\n return self.submitAsync(message,bindings=bindings,request_options=request_options).result()\n",line 148,in submitAsync\n return conn.write(message)\n"," File \"/var/task/gremlin_python/driver/connection.py\",line 55,in write\n self.connect()\n",line 45,in connect\n self._transport.connect(self._url,self._headers)\n"," File \"/var/task/gremlin_python/driver/tornado/transport.py\",line 40,in connect\n self._ws = self._loop.run_sync(\n"," File \"/var/task/tornado/ioloop.py\",line 576,in run_sync\n return future_cell[0].result()\n"
]
解决方法
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