我正在尝试读取具有标签的 YAML 文件!vault在里面。我收到错误：
无法确定标签“!vault”的构造函数
在阅读了几篇博客后，我了解到我需要指定一些构造函数来解决此问题，但我不清楚如何做到这一点。
import yaml
from yaml.loader import SafeLoader
with open('test.yml' ) as stream:
try:
inventory_info = yaml.safe_load(stream)
except yaml.YAMLError as exc:
print(exc)
User = inventory_info['all']['children']['linux']['vars']['user']
key = inventory_info['all']['children']['linux']['vars']['key_file']
我正在使用的 YAML 文件：
all:
children:
rhel:
hosts: 172.18.144.98
centos:
hosts: 172.18.144.98
linux:
children:
rhel:
centos:
vars:
user: "o9ansibleuser"
key_file: "test.pem"
ansible_password: !vault |
$ANSIBLE_VAULT;2.1;AES256
3234567899353936376166353
要么使用from_yaml实用功能：
from ansible.parsing.utils.yaml import from_yaml
# inventory_info = yaml.safe_load(stream) # Change this
inventory_info = from_yaml(stream) # to this
或者将构造函数添加到yaml.SafeLoader:
from ansible.parsing.yaml.objects import AnsibleVaultEncryptedUnicode
def construct_vault_encrypted_unicode(loader, node):
value = loader.construct_scalar(node)
return AnsibleVaultEncryptedUnicode(value)
yaml.SafeLoader.add_constructor(u'!vault', construct_vault_encrypted_unicode)
inventory_info = yaml.safe_load(stream)
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