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

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. If you would instead like to pass it as a query parameter, please take a look here. — Chris
– Chris, Commented Aug 27, 2022 at 4:31

elaadani · Accepted Answer · 2022-08-27 01:51:32Z

0

After a little research, I have a solution:


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

answered Aug 27, 2022 at 1:51

elaadani

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