How can I pass a URL as a parameter to GET endpoint in FastAPI?

Question

For example, I have this URL

https://store.epicgames.com/es-ES/p/dead-island-2--gold-edition

@app.get("/Datos/{CI}")
async def read_CI(CI: str):
    return{"CI_Datos":CI}```

This is my curl

curl -X 'GET' \
'http://127.0.0.1:8000/Datos/%2F%2F' \
-H 'accept: application/json'

and I want the API to be able to get that URL.

I get error 404 when I pass a URL. I understand that the problem is the /.

If you are looking for how to pass a URL as a path parameter, please have a look at [this answer](https://stackoverflow.com/a/72815364/17865804). If you would instead like to pass it as a query parameter, please take a look [here](https://stackoverflow.com/a/72068032/17865804). — Chris, Aug 27 '22 at 04:31

score 1 · Answer 1 · answered Aug 27 '22 at 01:51

1

After a little research, I have a solution:


@app.get("/Datos")
async def get_url(
    url: str = None
):
    return {"url": url}

answered Aug 27 '22 at 01:51

elaadani

1 Answers1