PyYAML 错误：无法确定标签“！vault”的构造函数

Question

我正在尝试读取包含标签 !vault 的 YAML 文件。我得到了错误：

无法确定标签“!vault”的构造函数

看了几篇博客，我明白我需要指定一些构造函数来解决这个问题，但我不清楚如何去做。

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

Answer 1

要么使用from_yaml 实用函数：

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)