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如何将以下命令放入 app.yaml 文件中。我想知道我做错了什么?

uwsgi --socket 127.0.0.1:3000 --file /home/james/james-api/runserver.py --callable app --processes 2

这是下面的 yaml 文件:

uwsgi: 
  socket: 127.0.0.1:3000
  python-path: /home/james/james-api/
  callable: runserver:app
  processes: 2
  pidfile: /tmp/uwsgi.pid
  daemonize: /var/log/uwsgi.log
  master: 1
  workers: 2
  chmod-socket: 666
  auto-procname: 1
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1 回答 1

0

我所要做的就是删除--

uwsgi: 
  socket: 127.0.0.1:3000 
  file: /home/james/james-api/runserver.py
  python-path: .
  callable: app 
  processes: 2
  master: 1
  workers: 2
  chmod-socket: 666
  auto-procname: 1
  pidfile: /tmp/uwsgi.pid
  daemonize: /var/log/uwsgi.log
于 2013-03-31T15:54:29.290 回答