FastAPI post does not recognize my parameter

Question

I am usually using Tornado, and trying to migrate to FastAPI.

Let's say, I have a very basic API as follows:

@app.post("/add_data")
async def add_data(data):
    return data

When I am running the following Curl request: curl http://127.0.0.1:8000/add_data -d 'data=Hello'

I am getting the following error:

{"detail":[{"loc":["query","data"],"msg":"field required","type":"value_error.missing"}]}

So I am sure I am missing something very basic, but I do not know what that might be.

Future readers might find [this answer](https://stackoverflow.com/a/70636163/17865804) and [this answer](https://stackoverflow.com/a/70640522/17865804) helpful as well. — Chris, Apr 27 '23 at 05:21

JPG · Accepted Answer · 2020-09-25T04:00:29.937

8

Since you are sending a string data, you have to specify that in the router function with typing as

from pydantic import BaseModel


class Payload(BaseModel):
    data: str = ""


@app.post("/add_data")
async def add_data(payload: Payload = None):
    return payload

Example cURL request will be in the form,

curl -X POST "http://0.0.0.0:6022/add_data"  -d '{"data":"Hello"}'

edited Sep 25 '20 at 04:00

answered Sep 25 '20 at 03:51

JPG

82,442
19
127
206

Thank you it works! I am however a bit confused why I need to define a BaseModel for post API but not get. – Djoby Sep 25 '20 at 14:34
1

In your get request, the data are being sent as ***query parameters*** whereas here as ***payload*** – JPG Sep 25 '20 at 15:18
And you don't have to hard-code input parameters (such as `data` field in `Payload`), see how to pass and receive any dict or JSON, without having to predefine its keys: https://stackoverflow.com/a/70879659/9962007 – mirekphd Dec 12 '22 at 12:08

score 1 · Answer 2 · answered Sep 25 '20 at 10:27

In your case, you passing a form data to your endpoint. To process it, you need to install python-multipart via pip and rewrite your function a little:

from fastapi import FastAPI, Form

app = FastAPI()

@app.post('/add_data')
async def process_message(data: str = Form(...)):
    return data

If you need json data, check Arakkal Abu's answer.

score 0 · Answer 3 · answered Jan 08 '23 at 21:35

FastAPI docs advise this solution for strings in body:

just add Body(), like data: str = Body().

Complete example:

from fastapi import Body

@app.post("/add_data")
async def add_data(data: str = Body()):
    return data

That tells FastAPI that the value is expected in body, not in the URL.

FastAPI post does not recognize my parameter

3 Answers3