How to write in Django a custom management command which takes a URL as a parameter?

Question

I am learning to write custom management commands in Django. I would like to write a command which would take a given URL as a parameter. Something like:

python manage.py command http://example.com

I've read a documentation, but it is not clear to me how to do this. But I can write a command saying 'Hello World!';)

Answer 1

try this:

create a file under yourapp/management/commands/yourcommand.py with the following content:

from django.core.management.base import BaseCommand

class Command(BaseCommand):
    help = 'A description of your command'

    def add_arguments(self, parser):
        parser.add_argument(
            '--url', dest='url', required=True,
            help='the url to process',
        )

    def handle(self, *args, **options):
        url = options['url']
        # process the url

then you can call your command with

python manage.py yourcommand --url http://example.com

and either:

python manage.py --help

or

python manage.py yourcommand --help

will show the description of your command and the argument.

if you don't want to name the argument (the --url part), like in your example, just read the url('s) form args:

def handle(self, *args, **kwargs):
    for url in args:
        # process the url

hope this helps.