Python AWS |模块中的语法错误:无效语法
我已经创建了python脚本,它作为zip文件上传到AWS Lambda函数中,其中捆绑了stompy库 python 2.7的日志:-Python AWS |模块中的语法错误:无效语法,python,python-2.7,stomp,amazon-mq,Python,Python 2.7,Stomp,Amazon Mq,我已经创建了python脚本,它作为zip文件上传到AWS Lambda函数中,其中捆绑了stompy库 python 2.7的日志:- Response: null Request ID: "c334839f-ee46-11e8-8970-612f1dc92e41" Function Logs: START RequestId: c334839f-ee46-11e8-8970-612f1dc92e41 Version: $LATEST CONNECTION Started CONNECTI
Response:
null
Request ID:
"c334839f-ee46-11e8-8970-612f1dc92e41"
Function Logs:
START RequestId: c334839f-ee46-11e8-8970-612f1dc92e41 Version: $LATEST
CONNECTION Started
CONNECTION established
CONNECTION Subscribed
[WARNING] 2018-11-22T11:07:12.798Z c334839f-ee46-11e8-8970-612f1dc92e41 Unknown response frame type: '' (frame length was 3)
END RequestId: c334839f-ee46-11e8-8970-612f1dc92e41
REPORT RequestId: c334839f-ee46-11e8-8970-612f1dc92e41 Duration: 10027.75 ms Billed Duration: 10100 ms Memory Size: 128 MB Max Memory Used: 30 MB
我的代码:-
import time
import boto3
import stomp
kinesis_client = boto3.client('kinesis')
class Listener(stomp.ConnectionListener):
msg_list = []
def on_error(self, headers, message):
print('received an error "%s"' % message)
def on_message(self, headers, message):
print('received a message "%s"' % message)
kinesis_client.put_record(
StreamName='Purchasing',
Data=u'{}\r\n'.format(message).encode('utf-8'),
PartitionKey='0'
)
def lambda_handler(event, context):
conn = stomp.Connection(host_and_ports=[('b-4714-4441-8166-47aae158281a-1.mq.eu-central-1.amazonaws.com', 8162)])
lst = Listener()
conn.set_listener('Listener', Listener())
conn.start()
conn.connect(login='test_mq', passcode='test_mq')
conn.subscribe(destination='/queue/Purchasing', id='b-4714-4441-8166-47aae158281a', ack='auto')
message = lst.msg_list
print('Waiting for messages "%s"' % message)
time.sleep(10)
conn.disconnect()
return ''
我不确定为什么我的消息没有显示在我的输出中,相反,它总是显示响应:null EDIT:正如@Petesh所指出的,这个问题来自stompyexternal库,它还没有被移植到Python3
如果您检查源代码,可以找到以下内容:
except socket.timeout, exc:
这是python3的无效语法+
如果在python3.6/3.7环境中运行lambda,则语法无效
如果您选择python 2.7,这个问题可能会消失,但您还必须调整代码、库等。请将代码以文本形式发布,而不是屏幕截图。我们可能需要将您的代码复制并粘贴到我们自己的文本编辑器中,以重现您的问题,但我们无法从图形中执行此操作。@cdarke我已附上了代码。您可以澄清问题出在stompy模块上,该模块尚未移植到python3上-您的回答不清楚这是问题的触发因素。@AlexK。。非常感谢。如果选择Python2.7,我需要对代码做哪些调整。此外,如何或在何处获得python3+stompylibraries@AlexK.. 此外,我还更新了Python2.7的错误日志。这看起来像是python的另一种选择3@AlexK我已经使用了你的链接并下载了tar.gz文件。但是里面没有stomp.py文件