How to POST JSON data with Python Requests?

Question:

I need to POST a JSON from a client to a server. I’m using Python 2.7.1 and simplejson. The client is using Requests. The server is CherryPy. I can GET a hard-coded JSON from the server (code not shown), but when I try to POST a JSON to the server, I get “400 Bad Request”.

Here is my client code:

data = {'sender':   'Alice',
    'receiver': 'Bob',
    'message':  'We did it!'}
data_json = simplejson.dumps(data)
payload = {'json_payload': data_json}
r = requests.post("http://localhost:8080", data=payload)

Here is the server code.

class Root(object):

    def __init__(self, content):
        self.content = content
        print self.content  # this works

    exposed = True

    def GET(self):
        cherrypy.response.headers['Content-Type'] = 'application/json'
        return simplejson.dumps(self.content)

    def POST(self):
        self.content = simplejson.loads(cherrypy.request.body.read())

Any ideas?

Asked By: Charles R

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Answers:

It turns out I was missing the header information. The following works:

import requests

url = "http://localhost:8080"
data = {'sender': 'Alice', 'receiver': 'Bob', 'message': 'We did it!'}
headers = {'Content-type': 'application/json', 'Accept': 'text/plain'}
r = requests.post(url, data=json.dumps(data), headers=headers)
Answered By: Charles R

Starting with Requests version 2.4.2, you can use the json= parameter (which takes a dictionary) instead of data= (which takes a string) in the call:

>>> import requests
>>> r = requests.post('http://httpbin.org/post', json={"key": "value"})
>>> r.status_code
200
>>> r.json()
{'args': {},
 'data': '{"key": "value"}',
 'files': {},
 'form': {},
 'headers': {'Accept': '*/*',
             'Accept-Encoding': 'gzip, deflate',
             'Connection': 'close',
             'Content-Length': '16',
             'Content-Type': 'application/json',
             'Host': 'httpbin.org',
             'User-Agent': 'python-requests/2.4.3 CPython/3.4.0',
             'X-Request-Id': 'xx-xx-xx'},
 'json': {'key': 'value'},
 'origin': 'x.x.x.x',
 'url': 'http://httpbin.org/post'}
Answered By: Zeyang Lin

From requests 2.4.2 (https://pypi.python.org/pypi/requests), the “json” parameter is supported. No need to specify “Content-Type”. So the shorter version:

requests.post('http://httpbin.org/post', json={'test': 'cheers'})
Answered By: ZZY

Works perfectly with python 3.5+

client:

import requests
data = {'sender':   'Alice',
    'receiver': 'Bob',
    'message':  'We did it!'}
r = requests.post("http://localhost:8080", json={'json_payload': data})

server:

class Root(object):

    def __init__(self, content):
        self.content = content
        print self.content  # this works

    exposed = True

    def GET(self):
        cherrypy.response.headers['Content-Type'] = 'application/json'
        return simplejson.dumps(self.content)

    @cherrypy.tools.json_in()
    @cherrypy.tools.json_out()
    def POST(self):
        self.content = cherrypy.request.json
        return {'status': 'success', 'message': 'updated'}
Answered By: Ruhil Jaiswal

The better way is:

url = "http://xxx.xxxx.xx"
data = {
    "cardno": "6248889874650987",
    "systemIdentify": "s08",
    "sourceChannel": 12
}
resp = requests.post(url, json=data)
Answered By: ellen

Which parameter between data / json / files you need to use depends on a request header named Content-Type (you can check this through the developer tools of your browser).

When the Content-Type is application/x-www-form-urlencoded, use data=:

requests.post(url, data=json_obj)

When the Content-Type is application/json, you can either just use json= or use data= and set the Content-Type yourself:

requests.post(url, json=json_obj)
requests.post(url, data=jsonstr, headers={"Content-Type":"application/json"})

When the Content-Type is multipart/form-data, it’s used to upload files, so use files=:

requests.post(url, files=xxxx)
Answered By: xiaoming

It always recommended that we need to have the ability to read the JSON file and parse an object as a request body. We are not going to parse the raw data in the request so the following method will help you to resolve it.

def POST_request():
    with open("FILE PATH", "r") as data:
        JSON_Body = data.read()
    response = requests.post(url="URL", data=JSON_Body)
    assert response.status_code == 200
Answered By: Pushparaj

I solved it this way:

from flask import Flask, request
from flask_restful import Resource, Api


req = request.json
if not req :
    req = request.form
req['value']
Answered By: Birk Net
headers = {"charset": "utf-8", "Content-Type": "application/json"}
url = 'http://localhost:PORT_NUM/FILE.php'

r = requests.post(url, json=YOUR_JSON_DATA, headers=headers)
print(r.text)
Answered By: isa

With current requests you can pass in any data structure that dumps to valid JSON , with the json parameter, not just dictionaries (as falsely claimed by the answer by Zeyang Lin).

import requests
r = requests.post('http://httpbin.org/post', json=[1, 2, {"a": 3}])

this is particularly useful if you need to order elements in the response.

Answered By: Ash