如何公开查看我在amazon-ec2上托管的网站？

Question

我已经用ec2-12-34-567-89.us-west-2.compute.amazonaws.com.的DNS启动了一个Amazonec2实例。在那里，我设置了一个安全组，其中type=HTTP、protocol=TCP、端口range=80，source=0.0.0.0/。我登录到EC2实例并启动我的应用程序：

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

然后，通过浏览器，我尝试打开：http://ec2-12-34-567-89.us-west-2.compute.amazonaws.com并获得“无法连接”消息。

我在这里错过了什么？

编辑

对于端口80，其结果如下：

$ 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

显然，我需要以超级用户的身份运行我的代码。

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