How to convert a cURL request to Python request

Question

I have a cURL request that is working fine.

curl http://localhost:5000/models/images/generic/infer.json -XPOST -F job_id='123' -F dont_resize='dont_resize' -F snapshot_epoch='100' -F image_file='@/home/hellouser/Downloads/infer/Users/User01/Images/tiles/999/00.jpg'`

I have a python script where I want to execute the same request. But I get the following error,

{ "error": { "message": "'NoneType' object has no attribute 'iteritems'", "type": "AttributeError" } }

Here is the python code,

import requests
data = {
    'job_id': '123',
    'dont_resize': 'dont_resize',
    'snapshot_epoch': '100',
    'image_file': '@/home/hellouser/Downloads/infer/Users/User01/Images/tiles/999/00.jpg'    }

url = 'http://localhost:5000/models/images/generic/infer.json'
r = requests.post(url=url, data=data)

Any idea how to properly convert the code? Should I pass file=file in request?

You might find this site useful: https://curl.trillworks.com/ — fordy, Feb 04 '19 at 13:59
@fordy thanks, I already tried it. The output from there (trillworks) was still giving errors when I sent it from Python. — Mujeeb, Feb 04 '19 at 14:01
@Melon what seems to be missing from the code? I'll check if it needs to be added. — Mujeeb, Feb 04 '19 at 14:02
it looks like the path in your backend is complaining about something, so you'll need to post the piece of code that deals with `iteritems` also seems like you are working with `python 2.x` you might want to label it as such — gold_cy, Feb 04 '19 at 14:03
That's the problem...I don't (and can't) have access to the backend code. I've been doing hit and trials all day to no avail. — Mujeeb, Feb 04 '19 at 14:04
See https://stackoverflow.com/a/54143121/476 for starters… — deceze, Feb 04 '19 at 14:05

score 1 · Accepted Answer · answered Feb 04 '19 at 14:05

1

Try this:

data = {
    'job_id': '123',
    'dont_resize': 'dont_resize',
    'snapshot_epoch': '100',
}
files = {
    'image_file': open('/home/hellouser/Downloads/infer/Users/User01/Images/tiles/999/00.jpg', 'rb')
}

url = 'http://localhost:5000/models/images/generic/infer.json'
r = requests.post(url=url, data=data, files=files)

answered Feb 04 '19 at 14:05

Rafael Acorsi

109
3

Thanks. I did try this but it didn't work earlier (don't remember the error). But much code has changed after that, I'll give it another try and let you know (later). – Mujeeb Feb 04 '19 at 14:08
Ok @Mujeeb, just remember, the param name is `files` – Rafael Acorsi Feb 04 '19 at 14:10
Shouldn't the file be opened with just `r`, instead of `rb` for Python 3? – Mujeeb Feb 05 '19 at 07:54
use rb for binary – Rafael Acorsi Feb 05 '19 at 23:29
Thanks. Found a detailed answer here, https://stackoverflow.com/a/9644285 – Mujeeb Feb 06 '19 at 05:18

How to convert a cURL request to Python request

1 Answers1