How to get current path in FastAPI with domain?

Question

I have a simple route as below that written in FastAPI,

from fastapi import FastAPI

app = FastAPI()


@app.get("/foo/bar/{rand_int}/foo-bar/")
async def main(rand_int: int):
    return {"path": f"https://some-domain.com/foo/bar/{rand_int}/foo-bar/?somethig=foo"}

How can I get the current path "programmatically" with,

domain (some-domain.com)
path (/foo/bar/{rand_int}/foo-bar/)
and query parameters (?somethig=foo)

score 32 · Accepted Answer · answered Dec 25 '21 at 03:26

32

We can use the Request.url-(starlette doc) API to get the various URL properties. To get the absolute URL, we need to use the Request.url._url private API , as below

from fastapi import FastAPI, Request

app = FastAPI()


@app.get("/foo/bar/{rand_int}/foo-bar/")
async def main(rand_int: int, request: Request):
    return {"raw_url": request.url._url}

answered Dec 25 '21 at 03:26

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9

It's also possible to call `str(request.url)` to get an string of the url to avoid accessing a private variable. – Abbas Jul 17 '22 at 06:15
6

@Abbas Good point, much more pythonic. I use it in f-strings, e.g. `f"{request.url}: {some_log_message}"`. – Jason R Stevens CFA Oct 16 '22 at 01:09

How to get current path in FastAPI with domain?

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