50

I need to pass in an integer argument to a base command in Django.

For instance, if my code is:

from django.core.management import BaseCommand

class Command(BaseCommand):
    def handle(self, *args, **options, number_argument):
        square = number_argument ** 2
        print(square)

I want to run:

python manage.py square_command 4

so, it will return 16.

Is there a way I can pass an argument through the terminal to the command I want to run?

0

4 Answers 4

96

Add this method to your Command class:

def add_arguments(self, parser):
    parser.add_argument('my_int_argument', type=int)

You can then use your option in the code, like this:

def handle(self, *args, **options):
    my_int_argument = options['my_int_argument']

The benefit of doing it this way is that the help output is automatically generated for manage.py my_command --help

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6 Comments

As of Django 2.8, this is the way to do it.
How can you add default value?
what is the purpose of *args then?
it's a standard pattern in Python to add *args and **kwargs: it ensures that when people inherit from this class you still can add extra keyword args in the function you override.
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3

Yes. The mechanism for doing so is described here, but basically, you can get the argument from args[0].

2 Comments

Found a more complete answer here: stackoverflow.com/questions/10568864/…
@PeterGraham now you should use the add_arguments method, which was added in version 1.8, so I wouldn't recommend that more complete answer anymore.
2

This is pretty simple by implementing the add_arguments method. Then, you can get the value of this argument in the handle method as follows:

class Command(BaseCommand):
    help = 'Help for command'

    def add_arguments(self, parser):
        parser.add_argument('n', type=int, help='help for n')

    def handle(self, *args, **options):
        n = options['n']
        square = n ** 2
        print(square)

The parameter parser is an instance of argparse.ArgumentParser (see the docs). Now you can add as many arguments as you want by calling parser's add_argument method. In the code above, you are expecting a parameter n of type int which is gotten in the handle method from options.

If you want to see the help displayed, run the following:

python manage.py square_command --help

That will produce an output similar to the following one:

usage: manage.py create_documents [-h] [--version] [-v {0,1,2,3}] [--settings SETTINGS] [--pythonpath PYTHONPATH] [--traceback] [--no-color] [--force-color]
                                  [--skip-checks]
                                  n

Help for command

positional arguments:
  n                     help for n

Comments

1

django-typer allows this to be specified more concisely than the BaseCommand class which uses argparse. The above could be rewritten:

from django_typer import TyperCommand

class Command(TyperCommand):
    help = 'Help for command'

    def handle(self, n: int):
        print(n ** 2)

Comments

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