Double-nested router can't have empty path(despite having prefix at grandparent level)

[teuneboon]

Describe the bug
If I have a router setup like this: main router -> sub router -> sub sub router. And I include sub router with a prefix, I can't have an empty path parameter on any routes in the sub sub router.

To Reproduce
Steps to reproduce the behavior:

1. Create a file with double-nested routers:

from fastapi import APIRouter

sub_sub_router = APIRouter()
sub_router = APIRouter()
main_router = APIRouter()

main_router.include_router(sub_router, prefix='/foo')
sub_router.include_router(sub_sub_router)

2. Add a path operation function without a path:

@sub_sub_router.get('')
def test():
    return 'bar'

3. See error: Exception: Prefix and path cannot be both empty (path operation: test)

Expected behavior
Since at one point in the path there is a prefix, "test" should be served at /foo.

Additional context
This is probably a choice to make on whether you want to support this use-case. To explain why I do it like this: we have a pretty complex API, and to reduce complexity(and prevent files being thousands of lines long) each endpoint(i.e.: /users) has a folder structure like:

• users/
  • __init__.py <-- contains main users router, includes read & write.py's routers
  • read.py <-- contains a router with get/read actions for users
  • write.py <-- contains a router with write actions for users
• routes.py <-- includes the user router from users/__init__.py and sets the prefix to '/users'

So for my usecase I have 2 alternative ways to work around this issue:

• Include read/write directly in the routes.py(this removes some abstraction though, maybe at one point we'll split it even more than just read/write)
• Set the prefix in users/__init__.py to '/users'(this moves the route definitions to a bunch of different files which is kind of meh)

I'm perfectly fine with using one of these workarounds if I have to, but before I do I wanted to know if you see this behavior as a bug as well. I checked out pull request #415 to see if I could make a pull request that fixes this, but unfortunately I don't know how to reach the parent router in the code that was changed there.

[dmontagu]

@teuneboon

Here's a workaround that I think handles your issue pretty cleanly:

@sub_sub_router.get("$")
def test():
    return 'bar'

Why this works: the path passed to the route decorator here is actually used to build a regex. $ matches the end of a string, so "$" is basically equivalent to "", but the recently-modified router-inclusion-check just checks that the path is not empty (not that it starts with "/"). So it will not raise an error when you do the nested include. (Actually, starlette adds a $ to the path already when building the regex, but having a second $ doesn't change what matches.)

Note: this will prevent the check that the route starts with a "/" when you are done cascading your inclusions, so you'd need to take responsibility for that. But that seems like a pretty reasonable compromise given the design you are trying to achieve.

[teuneboon]

@dmontagu thanks, that should fix the issue for me.

[tiangolo]

Thanks for the help here @dmontagu ! 👏 🙇

Thanks for reporting back and closing the issue @teuneboon 👍

[HansBrende]

@teuneboon @tiangolo @dmontagu the generated swagger shows /route$ instead of /route for this approach... and calls the route that is suffixed with the dollar sign when you click "try it out". Is there a way to avoid this?

[tiangolo]

@HansBrende I'm not sure I understand what you mean. If you still have problems please create a new issue with a simple self-contained example that shows how to reproduce the error, that way we can help you.

[HansBrende]

@tiangolo I am referring to the approach suggested by dmontagu. It works, but the Open API (swagger) page displays the regex in the path and the "try it out" button calls the incorrect API path. Not sure what part of this you didn't understand, perhaps you can elaborate.

[funilrys]

Hello there,

I had the same issue as @HansBrende regarding the generated OpenAPI schemas (also mentioned as "swagger" in his comment).

@dmontagu's answer - with "$" is just the right way. But, for those wondering about how to get the trailing $ away from the schema paths description here is what I found out.

I took the time to search for "OpenAPI" in the documentation and ended in the Extending OpenAPI chapter. I leave to you the lecture of the chapter but if we take the final example - as of today:

from fastapi import FastAPI
from fastapi.openapi.utils import get_openapi

app = FastAPI()


@app.get("/items/")
async def read_items():
    return [{"name": "Foo"}]


def custom_openapi():
    if app.openapi_schema:
        return app.openapi_schema
    openapi_schema = get_openapi(
        title="Custom title",
        version="2.5.0",
        description="This is a very custom OpenAPI schema",
        routes=app.routes,
    )
    openapi_schema["info"]["x-logo"] = {
        "url": "https://fastapi.tiangolo.com/img/logo-margin/logo-teal.png"
    }
    app.openapi_schema = openapi_schema
    return app.openapi_schema


app.openapi = custom_openapi

A simple change of the custom_openapi() function as follow fix our issue:

def custom_openapi():
    if app.openapi_schema:
        return app.openapi_schema
    openapi_schema = get_openapi(
        title="Custom title",
        version="2.5.0",
        description="This is a very custom OpenAPI schema",
        routes=app.routes,
    )
    openapi_schema["info"]["x-logo"] = {
        "url": "https://fastapi.tiangolo.com/img/logo-margin/logo-teal.png"
    }


    ## Change starts here.
    for path in {x for x in app.openapi_schema["paths"] if x.endswith("$")}:
        app.openapi_schema["paths"][path[:-1]] = app.openapi_schema["paths"][path]
        del app.openapi_schema["paths"][path]
    ### Change ends here.

    app.openapi_schema = openapi_schema
    return app.openapi_schema

I suppose that this is the right way but @tiangolo may correct me here.

Thanks for everything @dmontagu @tiangolo.