CODEWITHSUNDEEP
pythoncurlpython-requests
artem
artem

Reputation: 16777

Upload file with python requests not working

Note: I've already read all related questions & answers and tried that solutions without success.

I'm trying to upload file to the server with the following code:

with open('test.mp4', 'rb') as f:
    r = requests.post(
        url,
        headers={
            'Content-Type': 'multipart/form-data',
        },
        data=f
    )

But request always fails with:

requests.exceptions.ConnectionError: ('Connection aborted.', BrokenPipeError(32, 'Broken pipe'))

I've also tried to send it as files, not data.

I'm sure that server works fine, because if I send the same file to the same URL with curl it works:

curl -vvv -i -X POST -H "Content-Type: multipart/form-data" -F "[email protected]"  "https://vu.mycdn.me/upload.do?someskippedparams"

What's wrong with my code?

Upvotes: 1

Views: 2250

Answers (2)

gelonida
gelonida

Reputation: 5630

Normally this should work

with open('test.mp4', 'rb') as f:
    r = requests.post(
        url,
        headers={
            'Content-Type': 'multipart/form-data',
        },
        files={ "data": f},
    )

But somehow it fails for your server. Providing file name and mime type explicitly seems to solve the problem.

fname = "test.mp4"

with open(fname, "rb") as f:
    r = requests.post(
        url,
        files=[
            ("data", (os.path.basename(fname), f, "video/mp4")),
            ]
    )

print(r.status_code)
print(r.text)

Upvotes: 1

Harish Vutukuri
Harish Vutukuri

Reputation: 1150

import requests

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

params = (
    ('someskippedparams', ''),
)

files = {
    'data': ('test.mp4', open('test.mp4', 'rb')),
}

response = requests.post('https://vu.mycdn.me/upload.do', headers=headers, params=params, files=files)

Upvotes: 0

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