使用manage.py startapp在另一个目录中创建应用 [英] startapp with manage.py to create app in another directory

问题描述

我的 Django 项目结构是：

/proj
  /frontend
  /server
    /proj
    /app1
    /app2
  manage.py

我如何运行 python manage.py startapp app_name ，以便我新创建的应用位于<$ c $之内c> /服务器目录？我尝试在 server 目录中运行 django-admin.py startapp应用名称创建应用，但是最终此错误：

How do I run python manage.py startapp app_name so that my newly created apps are within the /server directory? I tried running django-admin.py startapp appname within the server directory to create the app but I would end up with this error:

$ ./manage.py runserver
Traceback (most recent call last):
  File "./manage.py", line 10, in <module>
    execute_from_command_line(sys.argv)
  File "/Users/bli1/Development/Django/CL/cherngloong/cherngloong/lib/python2.7/site-packages/django/core/management/__init__.py", line 351, in execute_from_command_line
    utility.execute()
  File "/Users/bli1/Development/Django/CL/cherngloong/cherngloong/lib/python2.7/site-packages/django/core/management/__init__.py", line 343, in execute
    self.fetch_command(subcommand).run_from_argv(self.argv)
  File "/Users/bli1/Development/Django/CL/cherngloong/cherngloong/lib/python2.7/site-packages/django/core/management/__init__.py", line 177, in fetch_command
    commands = get_commands()
  File "/Users/bli1/Development/Django/CL/cherngloong/cherngloong/lib/python2.7/site-packages/django/utils/lru_cache.py", line 101, in wrapper
    result = user_function(*args, **kwds)
  File "/Users/bli1/Development/Django/CL/cherngloong/cherngloong/lib/python2.7/site-packages/django/core/management/__init__.py", line 72, in get_commands
    for app_config in reversed(list(apps.get_app_configs())):
  File "/Users/bli1/Development/Django/CL/cherngloong/cherngloong/lib/python2.7/site-packages/django/apps/registry.py", line 137, in get_app_configs
    self.check_apps_ready()
  File "/Users/bli1/Development/Django/CL/cherngloong/cherngloong/lib/python2.7/site-packages/django/apps/registry.py", line 124, in check_apps_ready
    raise AppRegistryNotReady("Apps aren't loaded yet.")
django.core.exceptions.AppRegistryNotReady: Apps aren't loaded yet.

推荐答案

您可以指定<$ appname 之后的c $ c> / server / appname 目录作为目的地

You can specify the path to /server/appname directory after appname as the destination i.e. where the Django app directory structure will be created.

从 startapp 文档：

startapp <app_label> [destination] # startapp command usage 

为Django应用创建目录结构
当前目录中的给定应用程序名称或给定目的地。

如果仅给出了应用程序名称，则该应用程序目录将在当前工作目录的
中创建。

If only the app name is given, the app directory will be created in the current working directory.

如果提供了可选目标位置，则Django将使用现有的
目录而不是创建新目录一个

If the optional destination is provided, Django will use that existing directory rather than creating a new one

因此，您可以指定 / server / appname 的路径目录作为目的地值。

So, you can specify the path to your /server/appname directory as the destination value.

django-admin.py startapp appname [destination] # specify destination

您需要做什么？

1。。首先需要在内创建目录 appname / server 。

1. You need to first create a directory appname inside /server.

mkdir /server/appname # create directory from root level

2。。然后，运行 startapp 命令来创建应用。

2. Then, run the startapp command to create the app.

django-admin.py startapp appname ./server/appname

这篇关于使用manage.py startapp在另一个目录中创建应用的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！