How to send http request with multiple data arguments

Question

With flask_server.py running:

from flask import Flask, request, Response
import logging

logging.basicConfig(level=logging.DEBUG, 
                    format='%(levelname)s-%(message)s')

app = Flask(__name__)


@app.route('/test', methods=['GET','POST'])
def route():   
    logging.info('get_json: %s : %s' % (request.get_json(), type(request.get_json())))
    logging.info('files: %s : %s' % (request.files, type(request.files)))
    return Response()

if __name__ == '__main__':
    app.run('0.0.0.0', 5000)

send the http request using requests.post method supplying it with Python dictionary as json argument:

import json, requests    
dictionary = {"file": {"url": "https://bootstrap.pypa.io/get-pip.py"}}  
response = requests.post("http://127.0.0.1:5000/test", json=dictionary)

Flask server logs that it gets the dictionary with Flask.request.get_json method:

root - INFO - get_json: {u'file': {u'url': u'https://bootstrap.pypa.io/get-pip.py'}} : <type 'dict'>
root - INFO - files: ImmutableMultiDict([]) : <class 'werkzeug.datastructures.ImmutableMultiDict'>

Again, via requests.post method's files argument send an open file object. Flask server will get it via Flask.request.files attribute:

files = {'key_1': open('/any_file.txt', 'rb')}
response = requests.post(url, files = files)

Flask server logs:

root - INFO - get_json: None : <type 'NoneType'>
root - INFO - files: ImmutableMultiDict([('file_slot_1', <FileStorage: u'any_file.txt' (None)>)]) : <class 'werkzeug.datastructures.ImmutableMultiDict'>

Lastly, send both: the dictionary and open file object using same requests.post method:

response = requests.post("http://127.0.0.1:5000/test",
                         json=dictionary, 
                         files=files)

Server logs that it gets the file but does not get the json dictionary.

Is it be possible to send a request supplying it with multiple data arguments: such as 'json' and 'files'?

aneroid · Accepted Answer · 2018-11-14 02:14:22Z

2

You can send multiple file objects using the file parameter and a list. However, your json data will need to be among one of those files. That's because even though you're just doing json=dictionary in the first request, it's actually sending Content-Type: application/json in the header. When you want to send multiple parts, you need to use multipart/form-data, via files=<a list>:

json_data = json.dumps({"file": {"url": "https://bootstrap.pypa.io/get-pip.py"}})
multiple_files = [
    ('targets', ('data.json', json_data, 'application/json')),
    ('targets', ('key_1', open('/any_file.txt', 'rb'), 'text/plain'))
]
response = requests.post(url, files=multiple_files)

The target's name here is assumed to be targets. You can make up your own if your Flask app is the recipient of the uploads.

answered Nov 14, 2018 at 2:14

aneroid

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