Convert this curl command to Python 3

Question

How can I send this curl out using Python? I did find similar requests, but cannot adapt my code to suite the query I have also tried using pycurl, following this example but without luck.

$ curl -v -X GET  -H "Accept: text/csv" -H "Authorization: Basic YW5kcmVhLmJvdHRpQHdzcC5jb206OWY5N2E5YTY2ZWU1MTMxZjdmNjk4MDcwZTFkODEwMjU0M2I0NTg1ZA==" "https://epc.opendatacommunities.org/api/v1/domestic/search"

Thanks

Possible duplicate of [adding header to python requests module](https://stackoverflow.com/questions/8685790/adding-header-to-python-requests-module) — Diptangsu Goswami, Sep 04 '19 at 10:45

score 2 · Accepted Answer · answered Sep 04 '19 at 10:42

If you are using the Python Requests package the following code snippet should work:

import requests

headers = {
    'Accept': 'text/csv',
    'Authorization': 'Basic YW5kcmVhLmJvdHRpQHdzcC5jb206OWY5N2E5YTY2ZWU1MTMxZjdmNjk4MDcwZTFkODEwMjU0M2I0NTg1ZA==',
}

response = requests.get('https://epc.opendatacommunities.org/api/v1/domestic/search', headers=headers)

response.status_code  # 200
response.text # "lmk-key,address1,address2,address3,postcode,buildi ..."

(Note: I used the https://curl.trillworks.com/ website to automatically make the conversion)

Convert this curl command to Python 3

1 Answers1