django management command: using LabelCommand - django

Question

i want to pass argument along with this management command. i run this code from command line as
python manage.py example1 amita
where example1 is name of my file and amita is argument. on running this i am getting error.i am pasting the traceback:
Traceback (most recent call last):
File "manage.py", line 79, in <module>
execute_manager(settings)
File "/usr/lib/python2.7/dist-packages/django/core/management/__init__.py", line 438, in execute_manager
utility.execute()
File "/usr/lib/python2.7/dist-packages/django/core/management/__init__.py", line 379, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
File "/usr/lib/python2.7/dist-packages/django/core/management/__init__.py", line 261, in fetch_command
klass = load_command_class(app_name, subcommand)
File "/usr/lib/python2.7/dist-packages/django/core/management/__init__.py", line 68, in load_command_class
return module.Command()
AttributeError: 'module' object has no attribute 'Command'
the code for example1.py is below
from django.core.management.base import LabelCommand
from django.core.management.base import BaseCommand
def hello(name):
print name
def hello1(name):
print name
class LabelCommand(BaseCommand):
"""
A management command which takes one or more arbitrary arguments
(labels) on the command line, and does something with each of
them.
Rather than implementing ``handle()``, subclasses must implement
``handle_label()``, which will be called once for each label.
If the arguments should be names of installed applications, use
``AppCommand`` instead.
"""
args = '<label label ...>'
label = 'label'
def handle(self, *labels, **options):
if not labels:
raise CommandError('Enter at least one %s.' % self.label)
output = []
for label in labels:
label_output = self.handle_label(label, **options)
if label_output:
output.append(label_output)
return '\n'.join(output)
def handle_label(self, label, **options):
"""
Perform the command's actions for ``label``, which will be the
string as given on the command line.
"""
hello(label)
hello1(label)
raise NotImplementedError()

Answer 1

Django already has a LabelCommand class that you should use:
from django.core.management.base import LabelCommand
Then you just have to override the handle_label command.