Ansible - 删除键值对

Question

我有一个字典列表

member:
  - name: test2
    orig: test2
  - name: test1
    orig: test1

并且想要从列表中的所有字典中删除orig键和值。

- name: Print Firewall Group Member
  debug:
    msg: "{{ item }}"
  loop: "{{ member }}"

TASK [Print Firewall Group Member] *****************************************************************************************
ok: [fortigate01] => (item={'name': 'test2', 'orig': 'test2'}) => {
    "msg": {
        "name": "test2",
        "orig": "test2"
    }
}
ok: [fortigate01] => (item={'name': 'test1', 'orig': 'test1'}) => {
    "msg": {
        "name": "test1",
        "orig": "test1"
    }
}

Answer 1

1. 使用过滤器ansible.utils.remove_keys 。例如，给定简化数据

member:
  - name: test2
    orig: test2
  - name: test1
    orig: test1

声明变量

names: "{{ member|ansible.utils.remove_keys(target=['orig']) }}"

给出

names:
  - name: test2
  - name: test1

(可选)使用过滤器 ansible.utils.keep_keys 。下面的声明给出了相同的结果

names: "{{ member|ansible.utils.keep_keys(target=['name']) }}"

下一个选项是json_query

names: "{{ member|json_query('[].{name: name}') }}"

---

用于测试的完整剧本示例

- hosts: localhost

  vars:

    member:
      - name: test2
        orig: test2
      - name: test1
        orig: test1

    names: "{{ member|ansible.utils.remove_keys(target=['orig']) }}"
    name2: "{{ member|ansible.utils.keep_keys(target=['name']) }}"
    name3: "{{ member|json_query('[].{name: name}') }}"

  tasks:

    - debug:
        var: names
    - debug:
        var: name2
    - debug:
        var: name3