将curl post转换为rest api to python

Question

我有这个curl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。

这是我到目前为止所得到的：

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'])

我要去找SyntaxError: invalid syntax

你能帮我把curl命令正确地“转换”成python吗？

Answer 1

使用Authorization header，您可以执行以下操作：

import requests, base64, json

url = 'https://satellite.example.com/katello/api/content_view_versions/incremental_update'

headers = {
    'Accept': 'application/json,version=2',
    'Content-Type': 'application/json',
    'Authorization': 'Basic {}'.format(base64.b64encode('admin:Password'))}

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

response = requests.post(url=url, data=json.dumps(data), headers=headers)

您也可以使用不带Authorization头的auth参数，如下所示

from requests.auth import HTTPBasicAuth

headers = {
    'Accept': 'application/json,version=2',
    'Content-Type': 'application/json'}

response = requests.post(url, auth=HTTPBasicAuth('admin', 'Password'), data=json.dumps(data), headers=headers)