Python 2.7 将curl示例转换为pycurl_Python 2.7_Curl_Pycurl_Postmark

Python 2.7 将curl示例转换为pycurl

python-2.7 curl

Python 2.7 将curl示例转换为pycurl,python-2.7,curl,pycurl,postmark,Python 2.7,Curl,Pycurl,Postmark,有人可以将下面的邮戳curl示例转换为pycurl吗 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@examp

有人可以将下面的邮戳curl示例转换为pycurl吗

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>'}"

curl-X POST”http://api.postmarkapp.com/email" \
-H“接受：应用程序/json”\
-H“内容类型：应用程序/json”\
-H“X-邮戳-Server-Token:ed742D75-5a45-49b6-a0a1-5b9ec3dc9e5d”\
-五\
-d{From:'sender@example.com，致：receiver@example.com'主题：'邮戳测试'，HtmlBody:'您好亲爱的邮戳用户。'

您可以使用类似的方法。这是一个基本的实现，但应该可以工作

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()

我可以，但我不会。显示你的代码，我会修改。这对我来说比较容易。非常简洁的例子。帮了我很多忙：）