将cURL命令转换为python请求

Question

I am new to cURL; I wanted to convert a curl command of this structure:

curl -X POST "http://127.0.0.1:8881/models/faewrfaw/v1/predict" -H "Content-Type:multipart/form-data" -F "data={\"key\": \"Path\"};type=application/json" -F "Path=@C:\Users\rtam\Music\1 2 3\TEST123.jpg"

to a python request. I used https://curl.trillworks.com/ to assist me and got this as the output:

import requests

headers = {
    'Content-Type': 'multipart/form-data',
}

files = {
    'data': (None, '{"key": "Path"};type'),
    'Path': ('C:\\Users\\rtam\\Music\\1 2 3\\TEST123.jpg', open('C:\\Users\\rtam\\Music\\1 2 3\\TEST123.jpg', 'rb')),
}

response = requests.post('http://127.0.0.1:8881/models/faewrfaw/v1/predict', headers=headers, files=files)

However, when testing the response in python I got a bad request/request the server did not understand. I noticed the trillworks website did not account for the type (application/json) in its formatting, does it need to be in the python script?

Answer 1

This answer might help you: How to send JSON as part of multipart POST-request

In your case it would be

files = {
    'Path': (
        'C:\\Users\\rtam\\Music\\1 2 3\\TEST123.jpg', 
        open('C:\\Users\\rtam\\Music\\1 2 3\\TEST123.jpg', 'rb'), 
        'application/octet-stream'
    )
    'data': (None, '{"key": "Path"}', 'application/json'),
}

response = requests.post(
    'http://127.0.0.1:8881/models/faewrfaw/v1/predict',
    files=files
)