How can I use python to execute a curl command

+1 vote

I want to execute a curl command in python.

Usually, I just need to enter the command in terminal and press return key. However, I don't know how it works in python.

The command shows below:

curl -d @request.json --header "Content-Type: application/json"

There is a request.json file to be sent to get response.

I searched a lot and got confused. I tried to write a piece of code, although I could not fully understand. It didn't work.

import pycurl
import StringIO

response = StringIO.StringIO()
c = pycurl.Curl()
c.setopt(c.URL, '')
c.setopt(c.WRITEFUNCTION, response.write)
c.setopt(c.HTTPHEADER, ['Content-Type: application/json','Accept-Charset: UTF-8'])
c.setopt(c.POSTFIELDS, '@request.json')
print response.getvalue()

The error message is 'Parse Error'.Can anyone tell me how to fix it? or how to get response from the sever correctly?

Oct 11, 2018 in Python by aryya
• 7,450 points

3 answers to this question.

+1 vote

For sake of simplicity, maybe you should consider using the standard library Requests.

An example with json response content would be something like:

import requests
r = requests.get('')

If you look for further information, in the Quickstart section, they have lots of working examples.


For your specific curl translation:

import requests
url = ''
payload = open("request.json")
headers = {'content-type': 'application/json', 'Accept-Charset': 'UTF-8'}
r =, data=payload, headers=headers)

Hope it works!!

If you are a beginner and need to know more about Python, It's recommended to go for Python Certification course today.


answered Oct 11, 2018 by charlie_brown
• 7,720 points

Its convert your curl command to python request
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0 votes

Curl is a linux command that allows a user to make requests to or from a network server. It is especially useful for making automated shell scripts for downloading files or posting data.


Call requests.get(url) and, data, headers) to make a curl request, or any web request. The url is the url of the specified endpoint, data is the payload to send, and headers should contain any relevant headers for the request.

url = ""
headers = {"content-type": "application/json", "Accept-Charset": "UTF-8"}

r =, data={"sample":"data"}, headers=headers)

post data to url

data = r.json()

store response



{'args': {}, 'data': 'sample=data', 'files': {}, 'form': {}, 'headers': {'x-forwarded-proto': 'https', 'host': '', 'content-length': '11', 'accept': '*/*', 'accept-charset': 'UTF-8', 'accept-encoding': 'gzip, deflate', 'content-type': 'application/json', 'user-agent': 'python-requests/2.22.0', 'x-forwarded-port': '443'}, 'json': None, 'url': ''}
answered Dec 14, 2020 by Gitika
• 65,910 points
0 votes

You could use urllib as @roippi said:

import urllib2
data = '{"nw_src": "", "nw_dst": "", "nw_proto": "ICMP", "actions": "ALLOW", "priority": "10"}'
url = 'http://localhost:8080/firewall/rules/0000000000000001'
req = urllib2.Request(url, data, {'Content-Type': 'application/json'})
f = urllib2.urlopen(req)
for x in f:
answered Dec 14, 2020 by Roshni
• 10,520 points

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