file upload works with curl and postman but not python requests

Question

I am attempting to upload a CSV file to an API (which gives me very little error info) using python requests library.

(I'm running Python 3.5 and using requests version 2.18.4 on OS X 10.11.6)

This curl command in terminal works fine: curl -F 'file=@/path/to/file.csv' myurl.com/upload -H "Authorization: TOKEN sometoken"

A multipart/form-data POST request from Postman also works, but I can't seem to make it work with the python requests library.

I've tried many variations of this request:

import requests
headers = {'Authorization': 'TOKEN sometoken', 'Content-Type': 'multipart/form-data'}

with open(file_path, 'rb') as f:
    r = requests.post(myurl, headers=headers, data=f)

## I've also tried data={"file": f}

I get a status code of 200, but the response is {"success": "false"} (frustratingly unhelpful).

What am I missing in the python request compared to the curl request?

EDIT: it seems that the -F flag for the curl command emulates an HTML form that has been submitted...is there a way to do this with requests?

t.m.adam · Accepted Answer · 2018-04-27 23:14:56Z

2

In your curl code you're using the -F parameter, which submits the data as a multipart message - in other words you're uploading a file.
With requests you can post files with the files parameter. An example:

import requests

headers = {'Authorization': 'TOKEN sometoken'}
data = {'file': open(file_path, 'rb')}
r = requests.post(myurl, headers=headers, files=data)

Note that requests creates the Content-Type header automatically, based on the submitted data.

answered Apr 27, 2018 at 23:14

t.m.adam

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2 Comments

dukeluke Over a year ago

This worked, thanks! I noticed the files param in the docs, but they only mention it specifically for multiple file uploads at the same time, which threw me off.

t.m.adam Over a year ago

That's the dvanced usage documentation (in case you want to upload multiple files with the same name), and it can be confusing. The example in quickstart in is much clearer.

Fire · Accepted Answer · 2018-04-27 22:34:29Z

1

The python open() function returns a file object. This is not what you want to send to the API.
Instead, you can use:

with open(file_path, 'rb') as f:
    r = requests.post(myurl, headers=headers, data=f.read())

Syntax taken from here.

answered Apr 27, 2018 at 22:34

Fire

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1 Comment

dukeluke Over a year ago

Unfortunately, that didn't do the trick. I wish I could provide more info, but I have absolutely no error info to go on.

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