我试图了解如何使用保险库加密单个变量.首先,我使用ansible-vault encrypt_string -n -p加密字符串,然后将输出写入我的playbook.当我执行playbook时,它说解密的字符串不是
JSON可序列化的.
加密字符串：“inline_name”
我也尝试使用inline_name和inlinename,每次都有相同的结果.
我的剧本：
---
- name: Build System
hosts: dev
tasks:
- name: Create
MysqL_db:
state: present
name: !vault |
$ANSIBLE_VAULT;1.1;AES256
39613261386438623937643062636166663638633062323939343734306334346537613233623064
3761633832326365356231633338396132646532313861350a316666376566616633376238313636
39343833306462323534623238333639663734626662623731666239366566643636386261643164
3861363730336331660a316165633232323732633364346636363764623639356562336536636136
6364
login_host: "{{ MysqL_host }}"
login_user: "{{ MysqL_user }}"
login_password: "{{ MysqL_pass }}"
- name: Check if can access plain text vars
debug:
msg: "{{ my_plain_txt }}"
错误信息：
An exception occurred during task execution. To see the full traceback,use -vvv. The error was: TypeError: u'"inline_name"' is not JSON serializable fatal: [127.0.0.1]: Failed! => {"Failed": true,"msg": "Unexpected failure during module execution.","stdout": ""}
解决方法
添加任务级变量：
- name: Create
MysqL_db:
state: present
name: "{{ MysqL_name }}"
login_host: "{{ MysqL_host }}"
login_user: "{{ MysqL_user }}"
login_password: "{{ MysqL_pass }}"
vars:
MysqL_name: !vault |
$ANSIBLE_VAULT;1.1;AES256
39613261386438623937643062636166663638633062323939343734306334346537613233623064
3761633832326365356231633338396132646532313861350a316666376566616633376238313636
39343833306462323534623238333639663734626662623731666239366566643636386261643164
3861363730336331660a316165633232323732633364346636363764623639356562336536636136
6364
