startapp with manage.py to create app in another directory
Question:
My Django project structure is:
/proj
/frontend
/server
/proj
/app1
/app2
manage.py
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.
Answers:
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.
From the startapp documentation:
startapp <app_label> [destination] # startapp command usage
Creates a Django app directory structure for the given app name in the
current directory or the given destination.
If only the app name is given, the app directory will be created in
the current working directory.
If the optional destination is provided, Django will use that existing
directory rather than creating a new one
So, you can specify the path to your ./server/appname directory as the destination value.
django-admin.py startapp appname [destination] # Specify destination
What do you need to do?
1. You need to first create a directory, appname, inside /server.
mkdir ./server/appname # Create a directory from the root level
2. Then, run the startapp command to create the app.
django-admin.py startapp appname ./server/appname
The simplest way I found is to go inside the server folder and create an __init__.py file:
touch __init__.py
Then, create any app you want:
django-admin startapp {{APP NAME}}
If you are already in the server directory, then you can run
python ../manage.py startapp appname
And appname will be created in the server directory instead of in the project root.
I always have my app in an internal folder (the same that Django creates, with the name of the project) following the design of Two Scoops of Django that is similar to what you want to do. When you want to create a new app, you can use, as the previous answer says,
python ../manage.py startapp my_new_app
from within the folder in which you want to create the app. Another thing, even easier that is what I do, is that you can run
django-admin startapp my_new_app
from this inner folder, of apps and it will work.
To Create a New App in Django project.
First:
- Go to your folder using Terminal, where you want to create app
Second:
- Type the below command in terminal,
django-admin startapp <new_app_name>
Now, you can check, new app created with the given name.
Use absolute path for the manage file work well for me, an example to run from the destination folder
python /home/user/project_name/manage.py startupapp app_name
If you started the project using cookiecutter-django, then steps are given in the link below.
The difference from the accepted answer and other answers I see here is that you need to change the name of the app in apps.py as an extra step.
https://github.com/pydanny/cookiecutter-django/issues/1725#issuecomment-407493176
To copy paste that here:
1 - create the <name-of-the-app> app with python manage.py startapp
2 - move <name-of-the-app> directory to <project_slug> directory
3 - edit <project_slug>/<name-of-the-app>/apps.py and
change name = "<name-of-the-app>" to name = "<project_slug>.<name-of-the-app>"
4 - add "<project_slug>.<name-of-the-app>.apps.<NameOfTheAppConfigClass>", on your LOCAL_APPS on config/settings/base.py
Here <project_slug> would be your "server" directory.
My Django project structure is:
/proj
/frontend
/server
/proj
/app1
/app2
manage.py
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.
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.
From the startapp documentation:
startapp <app_label> [destination] # startapp command usage
Creates a Django app directory structure for the given app name in the
current directory or the given destination.
If only the app name is given, the app directory will be created in
the current working directory.If the optional destination is provided, Django will use that existing
directory rather than creating a new one
So, you can specify the path to your ./server/appname directory as the destination value.
django-admin.py startapp appname [destination] # Specify destination
What do you need to do?
1. You need to first create a directory, appname, inside /server.
mkdir ./server/appname # Create a directory from the root level
2. Then, run the startapp command to create the app.
django-admin.py startapp appname ./server/appname
The simplest way I found is to go inside the server folder and create an __init__.py file:
touch __init__.py
Then, create any app you want:
django-admin startapp {{APP NAME}}
If you are already in the server directory, then you can run
python ../manage.py startapp appname
And appname will be created in the server directory instead of in the project root.
I always have my app in an internal folder (the same that Django creates, with the name of the project) following the design of Two Scoops of Django that is similar to what you want to do. When you want to create a new app, you can use, as the previous answer says,
python ../manage.py startapp my_new_app
from within the folder in which you want to create the app. Another thing, even easier that is what I do, is that you can run
django-admin startapp my_new_app
from this inner folder, of apps and it will work.
To Create a New App in Django project.
First:
- Go to your folder using Terminal, where you want to create app
Second:
- Type the below command in terminal,
django-admin startapp <new_app_name>
Now, you can check, new app created with the given name.
Use absolute path for the manage file work well for me, an example to run from the destination folder
python /home/user/project_name/manage.py startupapp app_name
If you started the project using cookiecutter-django, then steps are given in the link below.
The difference from the accepted answer and other answers I see here is that you need to change the name of the app in apps.py as an extra step.
https://github.com/pydanny/cookiecutter-django/issues/1725#issuecomment-407493176
To copy paste that here:
1 - create the <name-of-the-app> app with python manage.py startapp
2 - move <name-of-the-app> directory to <project_slug> directory
3 - edit <project_slug>/<name-of-the-app>/apps.py and
change name = "<name-of-the-app>" to name = "<project_slug>.<name-of-the-app>"
4 - add "<project_slug>.<name-of-the-app>.apps.<NameOfTheAppConfigClass>", on your LOCAL_APPS on config/settings/base.py
Here <project_slug> would be your "server" directory.