uwsgi with flask-script Manager











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I use docker-compose to tie every image for launch once.




  • nginx

  • python


In python, I used uwsgi to deploy flask web server.



[manage.py]



import unittest

from app.main import create_app, db
from app.main.model import user
from flask_script import Manager
from app import blueprint

app = create_app(os.getenv('BOILERPLATE_ENV') or 'dev')
app.register_blueprint(blueprint)

app.app_context().push()

manager = Manager(app)

@manager.command
def run():
app.run(host='0.0.0.0')

@manager.command
def test():
"""Runs the unit tests."""
tests = unittest.TestLoader().discover('app/test', pattern='test*.py')
result = unittest.TextTestRunner(verbosity=2).run(tests)
if result.wasSuccessful():
return 0
return 1

if __name__ == '__main__':
manager.run()


As you know, I use flask-script Manager.



So when I execute flaks server, command like this -> python3 manage.py run



[uwsgi.ini]



[uwsgi]
chdir = /home
http-socket = :5001
chmod-socket = 777
logto = /home/web.log
process = 2
wsgi-file = manage.py
callable = manager
daemonize = /home/uwsgi.log
lazy-apps = true
pyargv = run


I set wsgi-file = manage.py and callable = manager.



Also set pyargv = run to same effect with python3 manage.py run



But when I run server and access the web, it throws error.



TypeError: __call__() takes from 1 to 2 positional arguments but 3 were given



Full log here.



root@web_project_dev:/home# uwsgi --ini config/uwsgi.ini
[uWSGI] getting INI configuration from config/uwsgi.ini
root@web_project_dev:/home# tail -f uwsgi.log
*** WARNING: you are running uWSGI as root !!! (use the --uid flag) ***
your server socket listen backlog is limited to 100 connections
your mercy for graceful operations on workers is 60 seconds
mapped 72920 bytes (71 KB) for 1 cores
*** Operational MODE: single process ***
uWSGI running as root, you can use --uid/--gid/--chroot options
*** WARNING: you are running uWSGI as root !!! (use the --uid flag) ***
*** uWSGI is running in multiple interpreter mode ***
spawned uWSGI worker 1 (and the only) (pid: 65, cores: 1)
WSGI app 0 (mountpoint='') ready in 0 seconds on interpreter 0x55588beca210 pid: 65 (default app)
TypeError: __call__() takes from 1 to 2 positional arguments but 3 were given
[pid: 65|app: 0|req: 1/1] 192.168.192.4 () {40 vars in 746 bytes} [Tue Nov 13 05:20:44 2018] GET / => generated 0 bytes in 0 msecs (HTTP/1.0 500) 0 headers in 0 bytes (0 switches on core 0)


How can I insert command line parameter to uwsgi.ini?



Thanks.










share|improve this question


























    up vote
    0
    down vote

    favorite












    I use docker-compose to tie every image for launch once.




    • nginx

    • python


    In python, I used uwsgi to deploy flask web server.



    [manage.py]



    import unittest

    from app.main import create_app, db
    from app.main.model import user
    from flask_script import Manager
    from app import blueprint

    app = create_app(os.getenv('BOILERPLATE_ENV') or 'dev')
    app.register_blueprint(blueprint)

    app.app_context().push()

    manager = Manager(app)

    @manager.command
    def run():
    app.run(host='0.0.0.0')

    @manager.command
    def test():
    """Runs the unit tests."""
    tests = unittest.TestLoader().discover('app/test', pattern='test*.py')
    result = unittest.TextTestRunner(verbosity=2).run(tests)
    if result.wasSuccessful():
    return 0
    return 1

    if __name__ == '__main__':
    manager.run()


    As you know, I use flask-script Manager.



    So when I execute flaks server, command like this -> python3 manage.py run



    [uwsgi.ini]



    [uwsgi]
    chdir = /home
    http-socket = :5001
    chmod-socket = 777
    logto = /home/web.log
    process = 2
    wsgi-file = manage.py
    callable = manager
    daemonize = /home/uwsgi.log
    lazy-apps = true
    pyargv = run


    I set wsgi-file = manage.py and callable = manager.



    Also set pyargv = run to same effect with python3 manage.py run



    But when I run server and access the web, it throws error.



    TypeError: __call__() takes from 1 to 2 positional arguments but 3 were given



    Full log here.



    root@web_project_dev:/home# uwsgi --ini config/uwsgi.ini
    [uWSGI] getting INI configuration from config/uwsgi.ini
    root@web_project_dev:/home# tail -f uwsgi.log
    *** WARNING: you are running uWSGI as root !!! (use the --uid flag) ***
    your server socket listen backlog is limited to 100 connections
    your mercy for graceful operations on workers is 60 seconds
    mapped 72920 bytes (71 KB) for 1 cores
    *** Operational MODE: single process ***
    uWSGI running as root, you can use --uid/--gid/--chroot options
    *** WARNING: you are running uWSGI as root !!! (use the --uid flag) ***
    *** uWSGI is running in multiple interpreter mode ***
    spawned uWSGI worker 1 (and the only) (pid: 65, cores: 1)
    WSGI app 0 (mountpoint='') ready in 0 seconds on interpreter 0x55588beca210 pid: 65 (default app)
    TypeError: __call__() takes from 1 to 2 positional arguments but 3 were given
    [pid: 65|app: 0|req: 1/1] 192.168.192.4 () {40 vars in 746 bytes} [Tue Nov 13 05:20:44 2018] GET / => generated 0 bytes in 0 msecs (HTTP/1.0 500) 0 headers in 0 bytes (0 switches on core 0)


    How can I insert command line parameter to uwsgi.ini?



    Thanks.










    share|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I use docker-compose to tie every image for launch once.




      • nginx

      • python


      In python, I used uwsgi to deploy flask web server.



      [manage.py]



      import unittest

      from app.main import create_app, db
      from app.main.model import user
      from flask_script import Manager
      from app import blueprint

      app = create_app(os.getenv('BOILERPLATE_ENV') or 'dev')
      app.register_blueprint(blueprint)

      app.app_context().push()

      manager = Manager(app)

      @manager.command
      def run():
      app.run(host='0.0.0.0')

      @manager.command
      def test():
      """Runs the unit tests."""
      tests = unittest.TestLoader().discover('app/test', pattern='test*.py')
      result = unittest.TextTestRunner(verbosity=2).run(tests)
      if result.wasSuccessful():
      return 0
      return 1

      if __name__ == '__main__':
      manager.run()


      As you know, I use flask-script Manager.



      So when I execute flaks server, command like this -> python3 manage.py run



      [uwsgi.ini]



      [uwsgi]
      chdir = /home
      http-socket = :5001
      chmod-socket = 777
      logto = /home/web.log
      process = 2
      wsgi-file = manage.py
      callable = manager
      daemonize = /home/uwsgi.log
      lazy-apps = true
      pyargv = run


      I set wsgi-file = manage.py and callable = manager.



      Also set pyargv = run to same effect with python3 manage.py run



      But when I run server and access the web, it throws error.



      TypeError: __call__() takes from 1 to 2 positional arguments but 3 were given



      Full log here.



      root@web_project_dev:/home# uwsgi --ini config/uwsgi.ini
      [uWSGI] getting INI configuration from config/uwsgi.ini
      root@web_project_dev:/home# tail -f uwsgi.log
      *** WARNING: you are running uWSGI as root !!! (use the --uid flag) ***
      your server socket listen backlog is limited to 100 connections
      your mercy for graceful operations on workers is 60 seconds
      mapped 72920 bytes (71 KB) for 1 cores
      *** Operational MODE: single process ***
      uWSGI running as root, you can use --uid/--gid/--chroot options
      *** WARNING: you are running uWSGI as root !!! (use the --uid flag) ***
      *** uWSGI is running in multiple interpreter mode ***
      spawned uWSGI worker 1 (and the only) (pid: 65, cores: 1)
      WSGI app 0 (mountpoint='') ready in 0 seconds on interpreter 0x55588beca210 pid: 65 (default app)
      TypeError: __call__() takes from 1 to 2 positional arguments but 3 were given
      [pid: 65|app: 0|req: 1/1] 192.168.192.4 () {40 vars in 746 bytes} [Tue Nov 13 05:20:44 2018] GET / => generated 0 bytes in 0 msecs (HTTP/1.0 500) 0 headers in 0 bytes (0 switches on core 0)


      How can I insert command line parameter to uwsgi.ini?



      Thanks.










      share|improve this question













      I use docker-compose to tie every image for launch once.




      • nginx

      • python


      In python, I used uwsgi to deploy flask web server.



      [manage.py]



      import unittest

      from app.main import create_app, db
      from app.main.model import user
      from flask_script import Manager
      from app import blueprint

      app = create_app(os.getenv('BOILERPLATE_ENV') or 'dev')
      app.register_blueprint(blueprint)

      app.app_context().push()

      manager = Manager(app)

      @manager.command
      def run():
      app.run(host='0.0.0.0')

      @manager.command
      def test():
      """Runs the unit tests."""
      tests = unittest.TestLoader().discover('app/test', pattern='test*.py')
      result = unittest.TextTestRunner(verbosity=2).run(tests)
      if result.wasSuccessful():
      return 0
      return 1

      if __name__ == '__main__':
      manager.run()


      As you know, I use flask-script Manager.



      So when I execute flaks server, command like this -> python3 manage.py run



      [uwsgi.ini]



      [uwsgi]
      chdir = /home
      http-socket = :5001
      chmod-socket = 777
      logto = /home/web.log
      process = 2
      wsgi-file = manage.py
      callable = manager
      daemonize = /home/uwsgi.log
      lazy-apps = true
      pyargv = run


      I set wsgi-file = manage.py and callable = manager.



      Also set pyargv = run to same effect with python3 manage.py run



      But when I run server and access the web, it throws error.



      TypeError: __call__() takes from 1 to 2 positional arguments but 3 were given



      Full log here.



      root@web_project_dev:/home# uwsgi --ini config/uwsgi.ini
      [uWSGI] getting INI configuration from config/uwsgi.ini
      root@web_project_dev:/home# tail -f uwsgi.log
      *** WARNING: you are running uWSGI as root !!! (use the --uid flag) ***
      your server socket listen backlog is limited to 100 connections
      your mercy for graceful operations on workers is 60 seconds
      mapped 72920 bytes (71 KB) for 1 cores
      *** Operational MODE: single process ***
      uWSGI running as root, you can use --uid/--gid/--chroot options
      *** WARNING: you are running uWSGI as root !!! (use the --uid flag) ***
      *** uWSGI is running in multiple interpreter mode ***
      spawned uWSGI worker 1 (and the only) (pid: 65, cores: 1)
      WSGI app 0 (mountpoint='') ready in 0 seconds on interpreter 0x55588beca210 pid: 65 (default app)
      TypeError: __call__() takes from 1 to 2 positional arguments but 3 were given
      [pid: 65|app: 0|req: 1/1] 192.168.192.4 () {40 vars in 746 bytes} [Tue Nov 13 05:20:44 2018] GET / => generated 0 bytes in 0 msecs (HTTP/1.0 500) 0 headers in 0 bytes (0 switches on core 0)


      How can I insert command line parameter to uwsgi.ini?



      Thanks.







      python uwsgi flask-script






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 13 at 5:22









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