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Accepted
Kevin C
Asked: 2024-01-25 16:33:08 +0800 CST

确保值是其他值的子字符串

  • 772

假设以下变量：

cron_d:
  script:
    - src: xfa.sh.j2
      dest: /opt/xfb_abc.sh
    - src: xfa.sh.j2
      dest: /opt/xfa_abc.sh
  cron:
    - name: agbcc
      job: /opt/xfa_abc.sh
    - name: aaaagbcc
      job: /opt/me.sh

我的目标是至少有一个值是 的cron_d.script.dest子串cron_d.cron.job。

我目前有这段代码，但它仅完全匹配字符串，而不仅仅是子字符串。

- name: assert that each script name matches at least once with a job name
  assert:
    that: 
      - "'{{ item.dest }}' in job_list"
  vars:
    job_list: "{{ cron_d.cron | map(attribute='job') | list }}"
  loop: "{{ cron_d.script }}"

我想扩展它，以便 的值dest必须出现在 中job，但作业可能包含更多字符串。

ansible

1 个回答

  1. Best Answer

    这可以简化为问题：

    问： “ dest的每一项必须是jobs中至少一项的子串。”

      jobs: [aaa, bbb, ccc, abc]
  dest: [aa, bc, xx]

    A：创建以下字典

      results: |
    {% filter from_yaml %}
    {% for i in dest %}
    {{ i }}: [{% for j in jobs %}{{ i in j }},{% endfor %}]
    {% endfor %}
    {% endfilter %}

    给出

      results:
    aa: [true, false, false, false]
    bc: [false, false, false, true]
    xx: [false, false, false, false]

    测试一下

        - assert:
        that: item.value is any
        success_msg: "[OK]  {{ item.key }} is a substring of some jobs."
        fail_msg: "[ERR] {{ item.key }} is not a substring of any job."
      loop: "{{ results|dict2items }}"

    给出（删节）

      msg: '[OK]  aa is a substring of some jobs.'
  msg: '[OK]  bc is a substring of some jobs.'
  msg: '[ERR] xx is not a substring of any job.'

    用于测试的完整剧本示例

    - hosts: localhost

  vars:

    jobs: [aaa, bbb, ccc, abc]
    dest: [aa, bc, xx]

    results: |
      {% filter from_yaml %}
      {% for i in dest %}
      {{ i }}: [{% for j in jobs %}{{ i in j }},{% endfor %}]
      {% endfor %}
      {% endfilter %}

  tasks:

    - debug:
        var: results|to_yaml

    - assert:
        that: item.value is any
        success_msg: "[OK]  {{ item.key }} is a substring of some jobs."
        fail_msg: "[ERR] {{ item.key }} is not a substring of any job."
      loop: "{{ results|dict2items }}"
    • 2

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