如何在Ansible中根据list1的name匹配list2并返回对应条目？

实现期望结果1（完整匹配条目） #

通过selectattr过滤器结合map提取list1的name列表，筛选list2中匹配的完整字典条目：

- hosts: localhost
  vars:
    list1 :
      - name: game1
      - name: game3
    list2:
      - name: game1
        type: application
        version: 123
        notes:
          type: url
          name: URLNAME
          page: https://urlname.com
      - name: game2
        type: application
        version: 223
        notes:
          type: url
          name: URLNAME
          page: https://urlname.com
      - name: game3
        type: application
        version: 333
        notes:
          type: url
          name: URLNAME
          page: https://urlname.com
      - name: game4
        type: application
        version: 443
        notes:
          type: url
          name: URLNAME
          page: https://urlname.com
  tasks:
    - name: 生成目标name列表
      set_fact:
        target_names: "{{ list1 | map(attribute='name') | list }}"

    - name: 筛选list2中匹配的完整条目
      set_fact:
        results: "{{ list2 | selectattr('name', 'in', target_names) | list }}"

    - name: 输出结果1
      debug:
        var: results

执行后results变量会完全符合你给出的期望结果1，包含匹配条目的所有字段。

实现期望结果2（精简版条目） #

在筛选基础上，通过map和extract过滤器提取指定字段，生成精简版结果：

- hosts: localhost
  vars:
    list1 :
      - name: game1
      - name: game3
    list2:
      - name: game1
        type: application
        version: 123
        notes:
          type: url
          name: URLNAME
          page: https://urlname.com
      - name: game2
        type: application
        version: 223
        notes:
          type: url
          name: URLNAME
          page: https://urlname.com
      - name: game3
        type: application
        version: 333
        notes:
          type: url
          name: URLNAME
          page: https://urlname.com
      - name: game4
        type: application
        version: 443
        notes:
          type: url
          name: URLNAME
          page: https://urlname.com
  tasks:
    - name: 生成目标name列表
      set_fact:
        target_names: "{{ list1 | map(attribute='name') | list }}"

    - name: 筛选并提取指定字段生成精简结果
      set_fact:
        results_slim: >
          {{ list2 | selectattr('name', 'in', target_names)
             | map('extract', ['name', 'type', 'version'])
             | map('zip', ['name', 'type', 'version'])
             | map('dict')
             | list }}

    - name: 输出结果2
      debug:
        var: results_slim

这段代码会把匹配条目只保留name、type、version三个字段，符合期望结果2的要求。

内容的提问来源于stack exchange，提问作者zwerk