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Fixed #6654 -- Slightly refactored the way 'startproject' and 'startapp' check for existing Python modules. Thanks, i_i 

git-svn-id: http://code.djangoproject.com/svn/django/trunk@7652 bcc190cf-cafb-0310-a4f2-bffc1f526a37
This commit is contained in:
Adrian Holovaty 2008-06-16 03:56:48 +00:00
parent cc64abfc1d
commit a0a06d1a89
2 changed files with 16 additions and 9 deletions
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13
django/core/management/commands/startapp.py
View File

@ -3,8 +3,7 @@ import os
from django.core.management.base import copy_helper, CommandError, LabelCommand from django.core.management.base import copy_helper, CommandError, LabelCommand
class Command(LabelCommand): class Command(LabelCommand):
help = ("Creates a Django app directory structure for the given app name" help = "Creates a Django app directory structure for the given app name in the current directory."
" in the current directory.")
args = "[appname]" args = "[appname]"
label = 'application name' label = 'application name'
@ -16,6 +15,7 @@ class Command(LabelCommand):
def handle_label(self, app_name, directory=None, **options): def handle_label(self, app_name, directory=None, **options):
if directory is None: if directory is None:
directory = os.getcwd() directory = os.getcwd()
# Determine the project_name by using the basename of directory, # Determine the project_name by using the basename of directory,
# which should be the full path of the project directory (or the # which should be the full path of the project directory (or the
# current directory if no directory was passed). # current directory if no directory was passed).
@ -23,6 +23,15 @@ class Command(LabelCommand):
if app_name == project_name: if app_name == project_name:
raise CommandError("You cannot create an app with the same name" raise CommandError("You cannot create an app with the same name"
" (%r) as your project." % app_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) copy_helper(self.style, 'app', app_name, directory, project_name)
class ProjectCommand(Command): class ProjectCommand(Command):

12
django/core/management/commands/startproject.py
View File

@ -3,8 +3,6 @@ import os
import re import re
from random import choice from random import choice
INVALID_PROJECT_NAMES = ('django', 'site', 'test')
class Command(LabelCommand): class Command(LabelCommand):
help = "Creates a Django project directory structure for the given project name in the current directory." help = "Creates a Django project directory structure for the given project name in the current directory."
args = "[projectname]" args = "[projectname]"
@ -20,13 +18,13 @@ class Command(LabelCommand):
# the parent directory. # the parent directory.
directory = os.getcwd() directory = os.getcwd()
# Check that the project_name cannot be imported.
try: try:
proj_name = __import__(project_name) __import__(project_name)
if proj_name:
raise CommandError("%r conflicts with the name of an existing Python module and cannot be used as a project name. Please try another name." % project_name)
except ImportError: except ImportError:
if project_name in INVALID_PROJECT_NAMES: pass
raise CommandError("%r contains an invalid project name. Please try another name." % project_name) else:
raise CommandError("%r conflicts with the name of an existing Python module and cannot be used as a project name. Please try another name." % project_name)
copy_helper(self.style, 'project', project_name, directory) copy_helper(self.style, 'project', project_name, directory)