Ansible：搜索以特定格式结尾的字符串

Question

我正在尝试从

checksums.txt

文件中过滤以 cmctl_linux_arm64 结尾的特定字符串，请参阅示例：

ff5e6efc85421bc46975ee60cf71bec1de3f983f27a3ba8e56a0953e8c9b8a07  cmctl_linux_arm64
bb5b8db1da6b4f7f046ebbf0aced0623e1b93990c474beb525296e72076b9421  cmctl_linux_arm64.spdx.sbom
6f903b8554f7e608ba3d156878f3720480b5dd693400e0354f2c88b94e4b9de0  cmctl_linux_arm64.tar.gz

考虑以下任务：

    - name: Set file fact
      ansible.builtin.set_fact:
        release_file: cmctl_linux_arm64
      run_once: true

    - name: Set checksums fact
      ansible.builtin.uri:
        url: https://github.com/cert-manager/cmctl/releases/download/v2.1.0/checksums.txt
        return_content: true
      register: checksums
      run_once: true

    - name: Set sha256 checksum fact
      ansible.builtin.set_fact:
        sha256_checksum: "{{ item.split(' ')[0] }}"
      loop: "{{ checksums.content.split('\n') }}"
      when: release_file in item
      run_once: true

我不确定如何使用 Ansible Regex 搜索，以便定义事实变量的结尾

release_file

并跳过存在的额外 2 个文件校验和。

编辑：使用以下格式修复问题：

when: item | regex_search(release_file + '$')

相比之下，我不能使用

select

:

when: item | select('search', release_file + '$')

或

in

:

when: item in regex_search(release_file + '$')

我想知道条件的正确写法是什么。

Answer 1

翻转物品。然后，匹配字符串开头的模式

  result: "{{ checksums.content.split('\n') | map('reverse') |
              select('match', release_file|reverse) |
              map('reverse') | map('split') | map('first') }}"

给予

  result:
  - ff5e6efc85421bc46975ee60cf71bec1de3f983f27a3ba8e56a0953e8c9b8a07

---

用于测试的完整剧本示例

- hosts: localhost

  vars:

    release_file: cmctl_linux_arm64
    checksums:
      content: |-
        ff5e6efc8542  cmctl_linux_arm64
        bb5b8db1da6b  cmctl_linux_arm64.spdx.sbom
        6f903b8554f7  cmctl_linux_arm64.tar.gz

    result: "{{ checksums.content.split('\n') | map('reverse') |
                select('match', release_file|reverse) |
                map('reverse') | map('split') | map('first') }}"

  tasks:

    - debug:
        var: result