python 3 run bash command get syntax error

Question

I'm trying to run bash commands from python3 script and I get an error. Command:

#!/usr/bin/python3

import os

os.system('curl -k --header "Authorization: 3NKNRNNUrFQtu4YsER6" --header "Accept: application/json" --header "Content-Type: application/json" https://192.168.1.1/alerts/index/limit:1/page:1/sort:id/direction:desc.json |  jq -r ''{"request": {"alert": {"alert": .[0].alert, "new": "test"}}}'' > 1.json')

Error response:

jq: error: syntax error, unexpected $end (Unix shell quoting issues?) at 
<top-level>, line 1:
{request:        
(23) Failed writing body

Answer 1

There's no need to use curl and jq ; Python has libraries to handle both HTTP requests and JSON data. ( requests is a 3rd-party library; json is part of the standard library.)

import json
import requests

with open("1.json", "w") as fh:
    response = requests.get("https://192.168.1.1/alerts/index/limit:1/page:1/sort:id/direction:desc.json",
                            headers={"Accept": "application/json",
                                     "Content-Type": "application/json",
                                     "Authorization": "3NKNRNNUrFQtu4YsER6"
                                    }
                           ).json()
    json.dump(fh, {'request': {'alert': {'alert': response[0]['alert'], 'new': 'test'}}})

If you insist on using curl and jq , use the subprocess module instead of os.system .

p = subprocess.Popen(["curl", "-k",
                      "--header", "Authorization: 3NKNRNNUrFQtu4YsER6",
                      "--header", "Accept: application/json",
                      "--header", "Content-Type: application/json",
                      "https://192.168.1.1/alerts/index/limit:1/page:1/sort:id/direction:desc.json"
                     ], stdout=subprocess.PIPE)

with open("1.json", "w") as fh:
    subprocess.call(["jq", "-r", '{request: {alert: {alert: .[0].alert, new: "test"}}}'], 
                    stdin=p.stdout,
                    stdout=fh)