"将"convert" curl rest api命令转换为python"

Question

我有这个命令：

curl -X POST -s -k -u "admin:Password" -d '

{
"add_content": {
    "errata_ids": ["RHSA-2016:2124","RHBA-2016:2889"]
},
"content_view_version_environments": [{
    "content_view_version_id": 28
}]
}
' \

-H "Accept:application/json,version=2" \ -H "Content-Type:application/json" \ https://satellite.example.com/katello/api/content_view_versions/incremental_update

我需要把它转换成python

到目前为止，我得到的是：

def post_json(location, json_data):


result = requests.post(
    location,
    data=json_data,
    auth=(USERNAME, PASSWORD),
    verify=SSL_VERIFY,
    headers=POST_HEADERS)

return result.json()

json_data = {
    "add_content": {
        "errata_ids": ["RHSA-2016:2124","RHBA-2016:2889"]
    },
    "content_view_version_environments": [{
        "content_view_version_id": 301
    }]
}

push_errata = post_json(katello_api + "content_view_versions + "/incremental_update/" + "content_view_version_environments/" + "add_content['RHSA-2016:1912'])

我得到SyntaxError: invalid syntax

你能帮我把curl命令正确地“转换”成python吗？你知道吗

Answer 1

如果您只想在Python中运行它，一个选项是使用subprocess。你知道吗

import subprocess
subprocess.Popen(['curl','-X', '''list your command here'''])