使用REST API发布文件：从curl示例到Python代码

Question

I'm trying to code the upload of a file to a web service through a REST API in Python. The service's documentation shows a example using curl as client:

curl -X POST -H \
-H "Content-Type: multipart/form-data" \
-F "file=filename.ext" \
-F "property1=value1" \
-F "property2=value2" \
-F "property3=value3" \
https://domain/api/endpoint

The difficulty for me is that this syntax doesn't match multipart form-data examples I found, including the requests documentation. I tried this, which doesn't work (rejected by the API):

import requests

file_data = [
    ("file", "filename.ext"),
    ("property1", "value1"),
    ("property2", "value2"),
    ("property3", "value3"),
]

response = requests.post("https://domain/api/endpoint",
    headers={"Content-Type": "multipart/form-data"}, files=file_data)

With the error: "org.apache.commons.fileupload.FileUploadException: the request was rejected because no multipart boundary was found"

Can anybody help in transposing that curl example to proper Python code?

Thanks!

R.

Answer 1

OK, looks like the web services documentation is wrong, and metadata simply needs to be sent as parameters. Moreover, I found in another request that you shouldn't set the header. So I was starting from a wrong example.