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我在这一切方面都是一个新手。尝试对树莓派的另一个应用程序进行故障排除,一直遇到这个问题。我读过 async 现在是 Python 3.7 的保留字,不能更改 server.py 中的脚本。可能有一个简单的解决方案,但我太新手看不到它。任何指导将不胜感激。

Traceback (most recent call last):
  File "./run.py", line 3, in <module>
    from modules import socketio, app, cbpi
  File "/home/pi/craftbeerpi3/modules/__init__.py", line 5, in <module>
    from flask_socketio import SocketIO, emit
  File "/usr/local/lib/python3.7/dist-packages/flask_socketio/__init__.py", line 7, in <module>
    from socketio import socketio_manage
  File "/usr/local/lib/python3.7/dist-packages/socketio/__init__.py", line 1, in <module>
    from .middleware import Middleware
  File "/usr/local/lib/python3.7/dist-packages/socketio/middleware.py", line 1, in <module>
    import engineio
  File "/usr/local/lib/python3.7/dist-packages/engineio/__init__.py", line 2, in <module>
    from .server import Server
  File "/usr/local/lib/python3.7/dist-packages/engineio/server.py", line 89
    self.async = None
             ^
SyntaxError: invalid syntax
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1 回答 1

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您使用的是旧版本的 python-socketio 和 python-engineio 包。升级到更新的版本以解决此问题。

于 2021-01-24T23:58:04.957 回答