We’re familiar to put a python file inside a folder, and create a
__init__.pyfile under the same folder, then we can easily import the file by import the folder, as the folder is transformed to a python module. But if we don’t have the __init__.py, how can we import it?
Suppose that we have a Flask project, during the development of Flask, we need to use the function flask_ctx_get_request_id() in the file request_id.py from the repo https://github.com/Workable/flask-log-request-id.
Here is the current folder tree, there’s only one file flask.py:
D:\xiang\git\test\flask_project │ flask.py
I add the repo as a submodule:
> git submodule add https://github.com/Workable/flask-log-request-id.git
Then my folder tree is like this:
D:\xiang\git\test\flask_project │ .gitmodules │ flask.py │ └─flask-log-request-id │ ├─flask_log_request_id │ │ ctx_fetcher.py │ │ filters.py │ │ parser.py │ │ request_id.py │ │ __init__.py │ │ │ └─extras │ celery.py │ __init__.py │ └─(more files and folders and ignored ...)
In flask.py, I try to import the function by importing the folder flask-log-request-id:
# flask.py from flask-log-request-id.flask_log_request_id.request_id import flask_ctx_get_request_id
Test the import:
> python .\flask.py File ".\flask.py", line 1 from flask-log-request-id.flask_log_request_id.request_id import flask_ctx_get_request_id ^ SyntaxError: invalid syntax
flask-log-request-id folder is not importable because it doesn’t contain the _init.py file. I don’t want to manually create it, it has no sense here. The workaround is to use the sys.path variable.
6.1.2. The Module Search Path
When a module named spam is imported, the interpreter first searches for a built-in module with that name. If not found, it then searches for a file named spam.py in a list of directories given by the variable sys.path. sys.path is initialized from these locations:
- The directory containing the input script (or the current directory when no file is specified).
- PYTHONPATH (a list of directory names, with the same syntax as the shell variable PATH).
- The installation-dependent default.
On file systems which support symlinks, the directory containing the input script is calculated after the symlink is followed. In other words the directory containing the symlink is not added to the module search path.
After initialization, Python programs can modify sys.path. The directory containing the script being run is placed at the beginning of the search path, ahead of the standard library path. This means that scripts in that directory will be loaded instead of modules of the same name in the library directory. This is an error unless the replacement is intended. See section Standard Modules for more information.
As per the official doc, I could add the path of the
flask-log-request-id folder to the sys.path variable, than the module
flask_log_request_id will be directly searchable by python process.
# flask.py import sys sys.path.append(r"d:/xiang/git/test/flask_project/flask-log-request-id") from flask_log_request_id.request_id import flask_ctx_get_request_id
The import will still fail due to the file
d:/xiang/git/test/flask_project/flask-log-request-id\flask_log_request_id\__init__.py, resolving this issue is out of scope of this blog, the adding path to the
sys.path to find the
flask_log_request_id module works well.