Pyright fails to detect aliased module with submodules

[ianliu]

Describe the bug
When aliasing a module, like import foo as f, one can access submodules with f.sub in the code, considering the submodule was brought into scope with imoport foo.sub, but Pyright fails in two ways: it reports that import foo.sub is unused and errors on f.sub.whatever.

To Reproduce

First, create a module foo like so:

mkdir foo
touch foo/__init__.py
echo "class User: pass" > foo/user.py

Now the following code will fail in pyright:

import foo as f
import foo.user    # "foo.user" is not accessed
print(f.user.User) # "user" is not a known member of module

Pyright version
1.1.267

Expected behavior
I would expect the code to type-check.

Additional context
Tested on mypy, and it type-checks

[erictraut]

This is intended behavior. You are relying on implicit import side effects, and pyright intentionally does not model those because they lead to code fragility.

If you want your code to work with pyright, you will need to change it to avoid this pattern. Instead, you can do one of the following:

1. Use the unaliased fully-qualified name to access symbols in the submodule.
2. Use an as clause to create an explicit alias to the submodule, as in import foo.user as u and then u.User.

[davidism]

In Flask-SQLAlchemy, I want to do this with SQLAlchemy and its submodules. `sqlalchemy... is a long name to type everywhere, and there's so many objects used from there that I want to keep them all under the module namespace instead of importing them all locally.

import sqlalchemy as sa
import sqlalchemy.orm
import sqlalchemy.events
# etc.

sa.create_engine
sa.orm.relationship
sa.events.DDLEvents

If I add as sa to every import, I get the same "not a known member" errors, so option 2 doesn't seem to work.

Could you elaborate on what is implicit and fragile about this? Both aliasing imports and importing submodules to add to the module's namespace are documented features of the import system. Why does pyright raise the error when using the alias?

[Aran-Fey]

@davidism I believe you've misunderstood the suggestion - it says to alias the submodule, like import sqlalchemy.orm as sa_orm.

But you're right that there is nothing fragile about this. After import foo.user, foo.user is guaranteed to exist, and since f is an alias of foo, so is f.user. This issue should certainly not be closed.