Split the list view code up in three pieces
1. getListContents which returns the required contents dictionary
2. _list which returns the response for a specified list of contents
3. list which constructs just one content dict and passes it to _list
This way it is easier to do step 1 and 2 in other code than list(),
which makes it possible to display multiple list pages (by calling
getListContents multiple times and passing the result to _list).
import os
from django.core.management.base import copy_helper, CommandError, LabelCommand
class Command(LabelCommand):
help = "Creates a Django app directory structure for the given app name in the current directory."
args = "[appname]"
label = 'application name'
requires_model_validation = False
# Can't import settings during this command, because they haven't
# necessarily been created.
can_import_settings = False
def handle_label(self, app_name, directory=None, **options):
if directory is None:
directory = os.getcwd()
# Determine the project_name by using the basename of directory,
# which should be the full path of the project directory (or the
# current directory if no directory was passed).
project_name = os.path.basename(directory)
if app_name == project_name:
raise CommandError("You cannot create an app with the same name"
" (%r) as your project." % app_name)
# Check that the app_name cannot be imported.
try:
__import__(app_name)
except ImportError:
pass
else:
raise CommandError("%r conflicts with the name of an existing Python module and cannot be used as an app name. Please try another name." % app_name)
copy_helper(self.style, 'app', app_name, directory, project_name)
class ProjectCommand(Command):
help = ("Creates a Django app directory structure for the given app name"
" in this project's directory.")
def __init__(self, project_directory):
super(ProjectCommand, self).__init__()
self.project_directory = project_directory
def handle_label(self, app_name, **options):
super(ProjectCommand, self).handle_label(app_name, self.project_directory, **options)