Get root path of Flask application
Andrew Mclaughlin 
I'm working on a Flask extension from which I want to create a directory in the project's root path on the file system.
Suppose we have this directory structure
/project /app /tests /my_folder manage.pymy_folder should be created dynamically by the extension, which is a test utility and wraps the application under test in the /tests directory. However, I'm struggling to determine the project's root path within my extension.
For now, I am trying to guess the path from the run file:
def root_path(self): # Infer the root path from the run file in the project root (e.g. manage.py) fn = getattr(sys.modules['__main__'], '__file__') root_path = os.path.abspath(os.path.dirname(fn)) return root_pathThis obviously breaks as soon as the tests are run from within the IDE instead of the manage.py. I could simply infer the project's root relative to the app or tests directory, but I don't want to make any assumptions regarding the name or structure of these directories (since multiple apps might be hosted as subpackages in a single package).
I was wondering if there is a best practice for this type of problem or an undocumented method which the Flask object provides (such as get_root_path).
2 Answers
app.root_path contains the root path for the application. This is determined based on the name passed to Flask. Typically, you should use the instance path (app.instance_path) not the root path, as the instance path will not be within the package code.
filename = os.path.join(app.instance_path, 'my_folder', 'my_file.txt')app.root_path is the absolute path to the root directory containing your app code.
app.instance_path is the absolute path to the instance folder. os.path.dirname(app.instance_path) is the directory above the instance folder. During development, this is next to or the same as the root path, depending on your project layout.
