python - How to send JSON as part of multipart POST-request

7
votes

I have following POST-request form (simplified):

POST /target_page HTTP/1.1  
Host: server_IP:8080
Content-Type: multipart/form-data; boundary=AaaBbbCcc

--AaaBbbCcc
Content-Disposition: form-data; name="json" 
Content-Type: application/json

{ "param_1": "value_1", "param_2": "value_2"}

--AaaBbbCcc
Content-Disposition: form-data; name="file"; filename="..." 
Content-Type: application/octet-stream

<..file data..>
--AaaBbbCcc--

I try to send POST-request with requests:

import requests
import json

file = "C:\\Path\\To\\File\\file.zip"
url = 'http://server_IP:8080/target_page'


def send_request():
    headers = {'Content-type': 'multipart/form-data; boundary=AaaBbbCcc'}

    payload = { "param_1": "value_1", "param_2": "value_2"}

    r = requests.post(url, files={'json': (None, json.dumps(payload), 'application/json'), 'file': (open(file, 'rb'), 'application/octet-stream')}, headers=headers)

    print(r.content)

if __name__ == '__main__':
    send_request()

but it returns status 400 with following comment:

Required request part \'json\' is not present.
The request sent by the client was syntactically incorrect.

Please point on my mistake. What should I change to make it work?

2
pythonjsonhttp-postpython-requestsmultipartform-data
You need to indicate Content-Type: application/json - hungneox
@noctilux: not for a multipart post you don't. - Martijn Pieters♦
Do not set the Content-type header yourself, leave that to requests to generate. - Martijn Pieters♦
In stackoverflow.com/questions/19439961/… is said not to encode the json part as json as a workaround - ralf htp
If you are not fixed on pyrequests you can use libcurl and PycURL (pycurl.io/docs/latest). In his thread is a working example for multipart POST with json in cURL: stackoverflow.com/questions/29231926/… - ralf htp

2 Answers

16
votes

You are setting the header yourself, including a boundary. Don't do this; requests generates a boundary for you and sets it in the header, but if you already set the header then the resulting payload and the header will not match. Just drop you headers altogether:

def send_request():
    payload = {"param_1": "value_1", "param_2": "value_2"}
    files = {
         'json': (None, json.dumps(payload), 'application/json'),
         'file': (os.path.basename(file), open(file, 'rb'), 'application/octet-stream')
    }

    r = requests.post(url, files=files)
    print(r.content)

Note that I also gave the file part a filename (the base name of the file path`).

For more information on multi-part POST requests, see the advanced section of the documentation.

0
votes

In case if someone searches ready to use method to transform python dicts to multipart-form data structures here is a simple gist example to do such transformation:

{"some": ["balls", "toys"], "field": "value", "nested": {"objects": "here"}}
    ->
{"some[0]": "balls", "some[1]": "toys", "field": "value", "nested[objects]": "here"}

To send some data you may want to use the multipartify method from this gist like this:

import requests  # library for making requests

payload = {
    "person": {"name": "John", "age": "31"},
    "pets": ["Dog", "Parrot"],
    "special_mark": 42,
}  # Example payload

requests.post("https://example.com/", files=multipartify(payload))

To send same data along with any file (as OP wanted) you may simply add it like this:

converted_data = multipartify(payload)
converted_data["attachment[0]"] = ("file.png", b'binary-file', "image/png")

requests.post("https://example.com/", files=converted_data)

Note, that attachment is a name defined by server endpoint and may vary. Also attachment[0] indicates that it is first file in you request - this is also should be defined by API documentation.