Ansible - 如何通过遍历另一个列表从字典列表中提取特定键[编辑]答案

【问题标题】：Ansible - How to extract specific keys from a List of Dictionary by iterating through another List [edit]Ansible - 如何通过遍历另一个列表从字典列表中提取特定键[编辑]
【发布时间】：2023-01-26 05:11:35
【问题描述】：

我有一个服务器名称列表和一个环境中所有集群的字典列表。字典列表包含该集群中的相关服务器。例如

"full_cluster_dict": [
{
    "key": "cluster_a", 
    "value": [
        "ca_server1",
        "ca_server2",
        "ca_server3",
        "ca_server4",
        "ca_server5",
        "ca_server6",
        "ca_server7",
        "ca_server8"
    ]
},
{
    "key": "cluster_b", 
    "value": [
        "cb_server1",
        "cb_server2",
        "cb_server3"
    ]
},
{
    "key": "cluster_c",
    "value": [
        "cc_server1",
        "cc_server2",
        "cc_server3",
        "cc_server4"
    ]
}

和

"server_list": [
    "ca_server1",
    "cb_server2",
    "ca_server6"
]

我想创建一个较小的字典列表，仅显示包含来自 server_list 的服务器的集群。例如

"needed_cluster_dict": [
  {
     "key: "cluster_a",
     "value": [
        "ca_server1",
        "ca_server2",
        "ca_server3",
        "ca_server4",
        "ca_server5",
        "ca_server6",
        "ca_server7",
        "ca_server8"
    ]
 },
 {
     "key": "cluster_b",
     "value" : [
        "cb_server1",
        "cb_server2",
        "cb_server3"
    ]
 }
]

我尝试了以下

- name: extract only relevant clusters based on the list of servers
  ansible.builtin.set_fact:
    needed_cluster_dict: "{{ needed_cluster_dict|d({}) | combine({item: cluster_filter}) }}"
  with_items: "{{ server_list }}"
  vars:
    cluster_filter: "{{ sds_dict|dict2items|json_query(_query) }}"
    _query: '[?value.contains(@, `{{ item }}`)].value'`

但这只返回一个字典，其中上面的服务器名称是键，每个键包含一个服务器列表，例如

needed_cluster_list: {
  "ca_server1: [
    "ca_server1",
    "ca_server2",
    "ca_server3",
    "ca_server4"
  ],
  "ca_server2: [
    "ca_server1",
    "ca_server2",
    "ca_server3",
    "ca_server4"
  ],
  ...
  ...
}

对我最初问题的编辑表示歉意，但在询问之后，我发现循环字典列表比循环字典更容易。

【问题讨论】：

标签： ansible ansible-facts ansible-filter

【解决方案1】：

创建所需集群的列表。测试列表的交集

  needed_cluster_str: |
    [{% for k,v in full_cluster_dict.items() %}
    {% if v|intersect(server_list)|length > 0 %}
    {{ k }},
    {% endif %}
    {% endfor %}]
  needed_cluster: "{{ needed_cluster_str|from_yaml }}"

给

  needed_cluster:
  - cluster_a
  - cluster_b

提取所需的列表

  needed_cluster_lists: "{{ needed_cluster|map('extract', full_cluster_dict)|list }}"

给

  needed_cluster_lists:
  - - ca_server1
    - ca_server2
    - ca_server3
    - ca_server4
    - ca_server5
    - ca_server6
    - ca_server7
    - ca_server8
  - - cb_server1
    - cb_server2
    - cb_server3

创建字典

  needed_cluster_dict: "{{ dict(needed_cluster|zip(needed_cluster_lists)) }}"

给

  needed_cluster_dict:
    cluster_a:
    - ca_server1
    - ca_server2
    - ca_server3
    - ca_server4
    - ca_server5
    - ca_server6
    - ca_server7
    - ca_server8
    cluster_b:
    - cb_server1
    - cb_server2
    - cb_server3

^{用于测试的完整剧本示例
- hosts: localhost

vars:

server_list: [ca_server1, cb_server2, ca_server6]
full_cluster_dict:
cluster_a:
- ca_server1
- ca_server2
- ca_server3
- ca_server4
- ca_server5
- ca_server6
- ca_server7
- ca_server8
cluster_b:
- cb_server1
- cb_server2
- cb_server3
cluster_c:
- cc_server1
- cc_server2
- cc_server3
- cc_server4

needed_cluster_str: |
[{% for k,v in full_cluster_dict.items() %}
{% if v|intersect(server_list)|length > 0 %}
{{ k }},
{% endif %}
{% endfor %}]
needed_cluster: "{{ needed_cluster_str|from_yaml }}"
needed_cluster_lists: "{{ needed_cluster|map('extract', full_cluster_dict)|list }}"
needed_cluster_dict: "{{ dict(needed_cluster|zip(needed_cluster_lists)) }}"

tasks:

- debug:
var: needed_cluster
- debug:
var: needed_cluster_lists
- debug:
var: needed_cluster_dict}

【讨论】：