I have a Cookiecutter Django project in which I'd like to add a "SiteSettings" model to control constants. My first instinct was to run manage.py startapp site. However I received the following message:
CommandError: 'site' conflicts with the name of an existing Python module and cannot be used as an app name. Please try another name.
Upon closer inspection, it appears that Cookiecutter already created a sites module within contrib, but with no models.py file. So I decided to make a models.py with the SiteSettings model in this directory.
However when I run makemigrations it says: No changes detected in app 'sites'
How can I add my desired model to this module and get migrations working for this module?
You will run into countless problems if you follow this path. Just use another name and do manage.py startapp newname.
I am trying to run
manage.py to validate models (as: manage.py validate) in the Interactive Shell of a Django project (called djangonew) using Pydev
The PYTHONPATH is set to include /djangonew ... so import djangonew and then dir(djangonew) actually gives me a name as 'settings' in the subfolder /djangonew/djangonew
but at the command-line I am unable run manage.py (and even find it)
How do I solve this issue? Thanks much
right click on project, open menu django-> custom command
and type verify ( or other command after "manage.py")
I'm trying to deploy a small django app on heliohost following the instructions in this post How to deploy a Django site using a different version of Django on Heliohost and also the official ones. But when I try to browse the site I only get the index of my public_html folder. Does anyone have any idea what I am doing wrong?
I know this is old post but if you still trying django on heliohost, try cookiecutter template for heliohost.
$ pip install cookiecutter
$ cookiecutter https://github.com/rahul-gj/cookiecutter-helio.git
project_name [mysite]: ---> Type your project name here.
helio_user [user_name]: ---> your heliohost username.
The folder with project name will be created in your working directory.
Paste the content of the folder (folder + manage.py file) on heliohost public_html directory.
go to yourusername.heliohost.org/yourproject
You should see something like this:
https://krydos.heliohost.org/djangotest/
Explore the django on heliohost.
I am a newbie at Django and everytime I try to run
python panel/manage.py startapp %app% (panel is my project) it gives me the error:
Error: '%app%' conflicts with the name of an existing Python module and cannot be used as an app name. Please try another name.
Am I doing something wrong?
Surely companies or contacts or stats is not the name of an existing Python module?
This is a fun one - your project and your app need to have different names. You probably created a project, then tried to startapp with the same name.
I was confused as well, until I realized that a Django project is a container for applications; this sequence makes it a bit clearer:
# first create a Project (container).
django-admin.py startproject Project
# create multiple apps
cd Project
python manage.py startapp polls
python manage.py startapp hello
...
Perhaps you need to
cd panel
python manage.py startapp yourappname
I'm not sure running the command from a directory above your project will work properly.
I had the same issue because I was trying to "restart" my app after carrying out changes, but startapp is meant to be used once to create a new app. To view changes, syncronize app with DB with python manage.py migrate and restart the server with python manage.py runserver instead.
From the django-admin docs
(manage.py does essentially the same thing as django-admin)
startapp <app_label> [destination]
django-admin startapp
Creates a Django app directory structure for
the given app name in the current directory or the given destination.
By default the directory created contains a models.py file and other
app template files. (See the source for more details.) 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. You can use ‘.’ to denote
the current working directory.
For example:
django-admin startapp myapp /Users/jezdez/Code/myapp
This message is displayed if you run "startapp" twice with the same app name. As pointed out above by the OP it doesn't reload the app, it creates one.
You should choose different names for your project and app in Codes:
django-admin startproject **my_project**
python manage.py startapp **my_app**
You need to create the directory before using the commands. Suppose you want a polls app inside apps folder.
mkdir apps apps/polls
python manage.py startapp polls apps/polls
I guess maybe you have already created the app's dir in panel dir manually. The command 'startapp' is to create an app automatically. If you already have one there, it fails.
I reproduced the issue and there's actually something not working as I expected.
I wonder if we stumbled upon a Django's bug, or a limitation that I don't understand.
Having a project called "project" and an empty folder app/newapp
…I tried:
python manage.py startapp newapp apps/newapp
It returns:
CommandError: 'newapp' conflicts with the name of an existing Python module and cannot be used as an app name. Please try another name.
But if I target ANY other route in which the last folder is not called the same name as the app I'm starting, it works.
So I ended up doing:
python manage.py startapp newapp apps/main
Using Django 2.1.3.
if you want to make an empty directory that will contain your new app
project-dir
└── blog
├── __init__.py
├── ...
├── blog-ext #this empty dir that will contain the new app
└── views.py
so instead of typing :
python manage.py newapp blog/blog-ext
it should be :
django-admin startapp newapp blog/blog-ext
Try classic "mysite" or "myproject". You can delete it anytime you want, so if it will accepted, then all your privious ideas conflict with Python modules.
Edit: I tried all your ideas, there was no error for me. So, if you installed support libraries or modules for django, then some of them can contains such names.
this error is because of the name conflicts between the app name and project name.you had given same name for your app and project .your project and app need to be different name .if you had given the same name the above mentioned error will occur .
understand the difference between app and project
Projects vs. apps
What’s the difference between a project and an app? An app is a Web application that does something – e.g., a Weblog system, a database of public records or a simple poll app. A project is a collection of configuration and apps for a particular Web site. A project can contain multiple apps. An app can be in multiple projects.
first create the project.
then create the app.
NOTE: name for app and project should be different
first create a project with projectname
django-admin.py startproject Projectname .
Then create app with appname. (to create your app make sure you are in the same directory manage.py and type this command)
python manage.py startapp Appname
It's the process how I got my doubt clear.
First, I created a directory inside my project directory and put __init__.py, models.py, admin.py, apps.py & views.py.
Then I ran python manage.py runserver & It work well.
Then as suggested on that page I used startapp command. I got this error :
CommandError: 'ucportal' conflicts with the name of an existing Python
module and cannot be used as an app name. Please try another name.
After that I deleted that directory and ran startapp command with same name and it worked fine.
So 'startapp' command is to create an app automatically. If you already have one there, it fails.
Answer given by #DAG worked for me.
I ran into this issue while trying to set up a Wagtail project.
Before creating the app, I had created and activated a virtualenv (using virtualenvwrapper) with the same name: $APPNAME. When I then ran wagtail start $APPNAME, Django looks for naming conflics in the $PYTHONPATH which in this instance points to /Users/User/.virtualenvs.
Naturally, this results in a conflict as /Users/User/.virtualenvs/$APPNAME already exists.
None of these answers helped me. In the end I ended up creating an app with a different name and then just renaming the directory to the app name I wanted all along. Note that you also will need to change the class name in apps.py to match your app name.
Just Simply Use This command
for Django Project Creation
python -m django startproject name_of_django_Project
for Django App Creation
python -m django startapp App_name
I had the same issue when working with wagtail cms. I got this error even there is no such a created app. This occurs when there is an app already that has the same name you need to create inside the site-packages directory.
Once you get this error, you need to check the following directory,
C:\Users\{user}\AppData\Local\Programs\Python\Python38-32\Lib\site-packages
If there is a package with the name same you want to create then you need to remove that package. Also make sure to check that package is important or not before deleting.
The application directory should be created first.
Example: apps/practice
The command appears to be duplicated, but it is correct.
Example: python manage.py startapp practice ./apps1/practice
I'm a Django newbie, but fairly experienced at programming. I have a set of related applications that I'd like to group into a sub-application but can not figure out how to get manage.py to do this for me.
Ideally I'll end up with a structure like:
project/
app/
subapp1/
subapp2/
I've tried manage.py startapp app.subapp1 and manage.py startapp app/subapp1
but this tells me that / and . are invalid characters for app names.
I've tried changing into the app directory and running ../manage.py subapp1 but that makes supapp1 at the top level. NOTE, I'm not trying to directly make a stand-alone application. I'm trying to do all this from within a project.
You can still do this :
cd app
django-admin startapp subapp1
This will work (create the application basic structure), however app and subapp1 will still be considered as two unrelated applications in the sense that you have to add both of them to INSTALLED_APPS in your settings.
Does this answer your question ? Otherwise you should tell more about what you are trying to do.
According to Django documentation,
If the optional destination is provided, Django will use that existing directory rather than creating a new one. You can use ‘.’ to denote the current working directory.
For example:
django-admin startapp myapp /Users/jezdez/Code/myapp
So, you can do it by this method:
Create sub_app1 directory in app directory
python manage.py startapp sub_app1 app/sub_app1
Django doesn't support "subapplications" per se. If you want code to be collected in packages within an app then just create them yourself. Otherwise you're just asking for pain.
Its Simple
Step 1 - Create Project
django-admin startproject app
cd app
Step 2 - Create api folder
mkdir api
cd api
touch __init__.py
cd ..
Step 3 - Create nested apps
python manage.py startapp user ./api/user
python manage.py startapp post ./api/post
python manage.py startapp comment ./api/comment
Step 4 - Register nested apps
INSTALLED_APPS = [
...
'api.user',
'api.post',
'api.comment',
]
Step 5 - Change name of Apps
Update the name in apps.py files in all three apps
class UserConfig(AppConfig):
default_auto_field = 'django.db.models.BigAutoField'
name = 'api.user'
class PostConfig(AppConfig):
default_auto_field = 'django.db.models.BigAutoField'
name = 'api.post'
class CommentConfig(AppConfig):
default_auto_field = 'django.db.models.BigAutoField'
name = 'api.comment'
and We are done!
Note : Step 2 is important
Go to your apps folder. Try:
python ../manage.py startapp app_name
django-admin.py startapp myapp /Users/jezdez/Code/myapp
Reference: Django Admin documentation
In order to create a nested app in the Django project, you have to follow the following steps:
Create a subdirectory in the parent directory where you want to add a sub-app.
mkdir finances/expences
N.B: Here finances is the existing parent app and expenses will be the next nested sub-app.
If the subdirectory is created successfully then run the following command in the terminal.
python manage.py startapp expenses finances/expenses
Please check the expenses subdirectory is now looks like a Django app and will behave like that.
Don't forget to register this app in the settings file and change the name from the config file of the app expenses, like -
class ExpensesConfig(AppConfig):
default_auto_field = 'django.db.models.BigAutoField'
name = 'finances.expenses'
Here, the name was only 'expenses' but it has been changed to 'finances.expenses' otherwise no migrations will be applied.
Use this command in the terminal:
python manage.py startapp application_name
there is really no any differ if you use django-admin or manage.py in this case - both will create
https://docs.djangoproject.com/en/4.0/ref/django-admin/#django-admin-and-manage-py
In addition, manage.py is automatically created in each Django project. It does the same thing as django-admin but also sets the DJANGO_SETTINGS_MODULE environment variable so that it points to your project’s settings.py file.