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Google Cloud Buildpack custom source directory for Python app

I am experimenting with Google Cloud Platform buildpacks, specifically for Python. I started with the Sample Functions Framework Python example app, and got that running locally, with commands:
pack build --builder=gcr.io/buildpacks/builder sample-functions-framework-python
docker run -it -ePORT=8080 -p8080:8080 sample-functions-framework-python
Great, let's see if I can apply this concept on a legacy project (Python 3.7 if that matters).
The legacy project has a structure similar to:
.gitignore
source/
main.py
lib
helper.py
requirements.txt
tests/
<test files here>
The Dockerfile that came with this project packaged the source directory contents without the "source" directory, like this:
COPY lib/ /app/lib
COPY main.py /app
WORKDIR /app
... rest of Dockerfile here ...
Is there a way to package just the contents of the source directory using the buildpack?
I tried to add this config to the project.toml file:
[[build.env]]
name = "GOOGLE_FUNCTION_SOURCE"
value = "./source/main.py"
But the Python modules/imports aren't set up correctly for that, as I get this error:
File "/workspace/source/main.py", line 2, in <module>
from source.lib.helper import mymethod
ModuleNotFoundError: No module named 'source'
Putting both main.py and /lib into the project root dir would make this work, but I'm wondering if there is a better way.
Related question, is there a way to see what project files are being copied into the image by the buildpack? I tried using verbose logging but didn't see anything useful.
Update:
The python module error:
File "/workspace/source/main.py", line 2, in <module>
from source.lib.helper import mymethod
ModuleNotFoundError: No module named 'source'
was happening because I moved the lib dir into source in my test project, and when I did this, Intellij updated the import statement in main.py without me catching it. I fixed the import, then applied the solution listed below and it worked.

I had been searching the buildpack and Google cloud function documentation, but I discovered the option I need on the pack build documentation page: option --path.
This command only captures the source directory contents:
pack build --builder=gcr.io/buildpacks/builder --path source sample-functions-framework-python
If changing the path, the project.toml descriptor needs to be in that directory too (or specify with --descriptor on command line).

Django - app in subfolder

I've created new django app in a subdirectory using the command:
python manage.py startapp appName subFolder/appName
but if I try to add this app to INSTALLED_APPS at the end of the list I see the following error:
ImportError: No module named appName
Does anyone know what I am doing wrong?

You need to include the subfolder when you add the app to INSTALLED_APPS, for example:
'subFolder.appName',
or
'subfolder.appName.apps.AppNameConfig',

I tried different options but no one can solve that issue.
Finally, I found a solution.
Simply go to the subFolder/appName/app.py file and
replace this line name = 'appname' with name = 'subfolder.appname'
And then you can simply add to the installed apps list.
'subfolder.appname'

Can a Python script use a Django database while not being in the same folder?

I have a python application (which we'll call app) I'd like to have a web front for. The application has many files and it's important the folder tree structure stays the way it is.
I've built a Django project and put it in a folder named "Web" in the app's folder, so now the folder tree looks like so:
[Data]
[Resources]
[Web]
[WebFront]
normal django app files
[Web]
__init__.py
settings.py
urls.py
wsgi.py
__init__.py
manage.py
main.py
Here's the code on the app's main.py:
import os
import django
os.environ.setdefault("DJANGO_SETTINGS_MODULE", "Web.Web.settings")
django.setup()
This code causes an exception on the django.setup() line as (I think) django does not find the project modules: ImportError: No module named WebFront (WebFront is the name of the django app)
I suspect this is caused because django runs in the directory of python app, and therefore cannot find the folder WebFront - Which should actually be Web/WebFront
Can this be done? Or should I reverse the order and put the python app in the django app?
This is not a duplicate of the following questions as the folder nesting causes a different problem (I think)
Django 1.7 throws django.core.exceptions.AppRegistryNotReady: Models aren't loaded yet
Easiest way to write a Python program with access to Django database functionality
Using only the DB part of Django

You can locate your main.py script where you like. However, if it is outside of the Web folder, then you will have to add Web to the Python path, otherwise imports like import Webfront are going to fail.
import sys
sys.path.append('/path/to/Web/')
Once you have done that, you can change the DJANGO_SETTINGS_MODULE to
os.environ.setdefault("DJANGO_SETTINGS_MODULE", "Web.settings")

startapp with manage.py to create app in another 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
Output:
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

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.

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.

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.

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 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.
New apps created are located in the project root, not inside the project slug #1725
To copy paste that here:
create the <name-of-the-app> app with python manage.py startapp
move <name-of-the-app> directory to the <project_slug> directory
edit <project_slug>/<name-of-the-app>/apps.py and change name = "<name-of-the-app>" to name = "<project_slug>.<name-of-the-app>"
add "<project_slug>.<name-of-the-app>.apps.<NameOfTheAppConfigClass>", in your LOCAL_APPS in file config/settings/base.py
Here <project_slug> would be your "server" directory.

Use an absolute path for the manage file. It works well for me, and here is an example to run from the destination folder:
python /home/user/project_name/manage.py startupapp app_name

How do I get all files for my django app into one folder?

I currently have a Django app named reserve. In the folder named "reserve" I have most of the content of my app (views.py, urls.py, models.py, templates folder). However, I have a folder outside of "reserve" named "booking" that has only my settings.py. I tried consolidating by putting the settings.py in "booking" into "reserve" but I seem to be getting an error. Any advice on how to have only one folder with all contents?
The error I get is: ImportError: Could not import settings 'booking.settings' (Is it on sys.path?): No module named booking.settings
overall project folder
booking folder
settings.py
init.py
reserve folder (the app)
views.py
admin.py
models.py
...

what version of django are you using? what command are you running ? assuming that you are just trying to run a command using manage.py you could just chnage manage.py to reflect this if you are going to permanently keep this project structure
#!/usr/bin/env python
import os
import sys
if __name__ == "__main__":
os.environ.setdefault("DJANGO_SETTINGS_MODULE", "your_project.settings")
from django.core.management import execute_from_command_line
execute_from_command_line(sys.argv)
you should change os.environ.setdefault to the correct directory for your settings.py

I'm assuming you have manage.py at the overal project directory? I'm also assuming you want to just runserver.
It seems you moved the settings.py file outside the overal project directory. If you do this, you'll have to specify the settings file path when you do runserver, or you can also export the django settings module to point to your new settings path (using dot notation). In this case, I think it would be (in the command line):
export DJANGO_SETTINGS_MODULE=<overal-project-folder-name>.booking.settings
python manage.py runserver