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python 2.7 - Convert curl example to pycurl

Could someone convert the following PostMark curl example to pycurl?

curl -X POST "http://api.postmarkapp.com/email" 

-H "Accept: application/json" 

-H "Content-Type: application/json" 

-H "X-Postmark-Server-Token: ed742D75-5a45-49b6-a0a1-5b9ec3dc9e5d" 

-v 

-d "{From: 'sender@example.com', To: 'receiver@example.com', Subject: 'Postmark test', HtmlBody: '<html><body><strong>Hello</strong> dear Postmark user.</body></html>'}"
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by (71.8m points)

You can use something like this. It's a basic implementation but it should work.

import pycurl, json

postmark_url = 'https://api.postmarkapp.com/email'

data = json.dumps({"From": "user@example.com", "To": "receiver@example.com", "Subject": "Pycurl", "TextBody": "Some text"})

c = pycurl.Curl()
c.setopt(pycurl.URL, github_url)
c.setopt(pycurl.HTTPHEADER, ['X-Postmark-Server-Token: API_TOKEN_HERE','Accept: application/json'])
c.setopt(pycurl.POST, 1)
c.setopt(pycurl.POSTFIELDS, data)
c.perform()

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