Send and receive file using Python: FastAPI and requests [duplicate]

Question

I'm trying to upload a file to a FastAPI server using requests.

I've boiled the problem down to its simplest components.

The client using requests:

import requests

files = {'file': ('foo.txt', open('./foo.txt', 'rb'))}
response = requests.post('http://127.0.0.1:8000/file', files=files)
print(response)
print(response.json())

The server using fastapi:

from fastapi import FastAPI, File, UploadFile
import uvicorn

app = FastAPI()

@app.post('/file')
def _file_upload(my_file: UploadFile = File(...)):
    print(my_file)

if __name__ == "__main__":
    uvicorn.run("main:app", host="0.0.0.0", port=8000, log_level="debug")

Packages installed:

fastapi
python-multipart
uvicorn
requests

Client Output: <Response [422]> {'detail': [{'loc': ['query', 'my_file'], 'msg': 'field required', 'type': 'value_error.missing'}]}

Server Output: INFO: 127.0.0.1:37520 - "POST /file HTTP/1.1" 422 Unprocessable Entity

What am I missing here?

JPG · Accepted Answer · 2020-12-02 05:57:32Z

27

FastAPI expecting the file in the my_file field and you are sending it to the file field.

it should be as

import requests

url = "http://127.0.0.1:8000/file"
files = {'my_file': open('README.md', 'rb')}
res = requests.post(url, files=files)

Also, you don't need a tuple to manage the upload file (we're dealing with a simple upload, right?)

answered Dec 2, 2020 at 5:57

JPG

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Comments

AUserOfThings · Accepted Answer · 2020-12-02 05:50:37Z

1

Remove the 'foo.txt' from your request.

It should look like

files = {'file': open('./foo.txt', 'rb')}

answered Dec 2, 2020 at 5:50

AUserOfThings

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1 Comment

prometheanSun Over a year ago

the ('foo.txt', ... is to add a file-name to the post-request: requests.readthedocs.io/en/master/user/quickstart I tried it without, and get same result.

JIsam · Accepted Answer · 2022-06-07 21:17:10Z

0

based on the documentation: https://requests.readthedocs.io/en/master/user/quickstart/

the following code should work (worked for me):

image = {'file':("your_file_name.extension", open("your_file_name.extension", 'rb'))}

resp = requests.post(url="your url", files=image)

answered Jun 7, 2022 at 21:17

JIsam

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