Creating django management commands

Question

I want to run a command which is something like this, but the function handle (self,*args,**options) doesn't seem to execute the nested functions.

How can I include my functions inside handle()?

from django.core.management.base import NoArgsCommand

class Command(NoArgsCommand):
    def handle(self, *args, **options):
        def hello():
            print "hello1"
        def hello1():
            print "hello2"

You must 'call' the functions after their definitions in order to execute them: hello(); hello1() — Don
– Don, Commented Apr 20, 2012 at 8:56

Don · Accepted Answer · 2012-04-20 08:55:20Z

3

You can also define functions 'on the fly':

class Command(NoArgsCommand):
    def handle_noargs(self):
        def hello():
            print "hello1"

        def hello1():
            print "hello2"

        hello()
        hello1()

or outside the command (as they where 'service' functions):

def hello():
    print "hello1"

def hello1():
    print "hello2"


class Command(NoArgsCommand):

    def handle_noargs(self):
        hello()
        hello1()

answered Apr 20, 2012 at 8:55

Don

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2 Comments

Amita Singh Over a year ago

:hey thanks a lot.. it really helped....one more doubt....if i want to call a function that takes an argument from command line then how i am i supposed to do that ????.Do i have to use class Command or some other class.

Don Over a year ago

Yes, you must extend BaseCommand and use 'handle' method. See here (docs.djangoproject.com/en/dev/howto/custom-management-commands) for two ways to have arguments: "*args" or "option_list"

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