如何公開查看我在Amazon EC2上托管的網站？

Question

我已經啟動了一個Amazon-ec2實例，並公開了ec2-12-34-567-89.us-west-2.compute.amazonaws.com的DNS。 我在那里設置了一個安全組，其中類型為HTTP，協議為TCP，端口范圍為80，源為0.0.0.0/。 我登錄到EC2實例並啟動我的應用程序：

$ python3 run.py
 * Running on http://0.0.0.0:0/

然后，使用瀏覽器嘗試打開： http : //ec2-12-34-567-89.us-west-2.compute.amazonaws.com並收到“無法連接”消息。

我在這里想念什么？

編輯

使用端口80，它最終會像這樣：

$ python3 run.py
 * Running on http://0.0.0.0:80/
Traceback (most recent call last):
  File "run.py", line 5, in <module>
    app.run(host="0.0.0.0", port=80)
  File "/opt/python/3.4.1/lib/python3.4/site-packages/flask/app.py", line 772, in run
    run_simple(host, port, self, **options)
  File "/opt/python/3.4.1/lib/python3.4/site-packages/werkzeug/serving.py", line 710, in run_simple
   inner()
  File "/opt/python/3.4.1/lib/python3.4/site-packages/werkzeug/serving.py", line 692, in inner
    passthrough_errors, ssl_context).serve_forever()
  File "/opt/python/3.4.1/lib/python3.4/site-packages/werkzeug/serving.py", line 486, in make_server
    passthrough_errors, ssl_context)
  File "/opt/python/3.4.1/lib/python3.4/site-packages/werkzeug/serving.py", line 410, in __init__
    HTTPServer.__init__(self, (host, int(port)), handler)
  File "/opt/python/3.4.1/lib/python3.4/socketserver.py", line 429, in __init__
    self.server_bind()
  File "/opt/python/3.4.1/lib/python3.4/http/server.py", line 133, in server_bind
    socketserver.TCPServer.server_bind(self)
  File "/opt/python/3.4.1/lib/python3.4/socketserver.py", line 440, in server_bind
    self.socket.bind(self.server_address)
PermissionError: [Errno 13] Permission denied

Answer 1

顯然我需要以超級用戶身份運行代碼。

$ **sudo** python3 run.py
 * Running on http://0.0.0.0:80/