Add confirmation step to custom Django management/manage.py command
<p>I created the following custom management command following <a href="https://docs.djangoproject.com/en/1.10/howto/custom-management-commands/" rel="nofollow">this tutorial</a>.</p>

<pre><code>from django.core.management.base import BaseCommand, CommandError
from django.contrib.auth.models import User

from topspots.models import Notification


class Command(BaseCommand):
    help = 'Sends message to all users'

    def add_arguments(self, parser):
        parser.add_argument('message', nargs='?')

    def handle(self, *args, **options):
        message = options['message']
        users = User.objects.all()
        for user in users:
            Notification.objects.create(message=message, recipient=user)

        self.stdout.write(
            self.style.SUCCESS(
                'Message:\n\n%s\n\nsent to %d users' % (message, len(users))
            )
        )
</code></pre>

<p>It works exactly as I want it to, but I would like to add a confirmation step so that before the <code>for user in users:</code> loop you are asked if you really want to send message X to N users, and the command is aborted if you choose "no". </p>

<p>I assume this can be easily done because it happens with some of the built-in management commands, but it doesn't seem to cover this in the tutorial and even after some searching and looking at the source for the built-in management commands, I have not been able to figure it out on my own.</p>

<p>You can use Python's <code>raw_input</code>/<code>input</code> function. Here's an example method from Django's <a href="https://github.com/django/django/blob/59afe61a970dd60df388e7cda9041ef3c0e770cb/django/db/migrations/questioner.py#L87" rel="nofollow">source code</a>:</p>

<pre><code>from django.utils.six.moves import input

def boolean_input(question, default=None):
    result = input("%s " % question)
    if not result and default is not None:
        return default
    while len(result) &lt; 1 or result[0].lower() not in "yn":
        result = input("Please answer yes or no: ")
    return result[0].lower() == "y"
</code></pre>

<p>Be sure to use the import from <code>django.utils.six.moves</code> if your code should be compatible with Python 2 and 3, or use <code>raw_input()</code> if you're on Python 2. <code>input()</code> on Python 2 will evaluate the input rather than converting it to a string. </p>

python django django-manage.py django-management-command