无法打印具有正则表达式匹配的变量，结果为Ansible

Question

下面是我的剧本，它搜索字符串“ SSLFile”并将匹配结果存储在名为“ target”的set_fact中

- name: Find certificate entries
  set_fact:
    target: "{{ filecontent.stdout | regex_findall('\\sSSLFile.*') }}"

- debug:
    msg: "{{ target }}"

上面的调试输出显示了三行匹配。参见下面的输出：

TASK [Find certificate entries] ***************************************
task path: /app/test.yml:908
ok: [10.9.9.11] => {
    "ansible_facts": {
        "target": [
            " SSLFile  /web/test1.crt", 
            " SSLFile  /web/SSL.crt", 
            " SSLFile  /web/test.crt"
        ]
    }, 
    "changed": false
}

我希望仅从文件名“ target”即第二列中获取文件名，即

/web/test1.crt
/web/SSL.crt
/web/test.crt

我尝试了以下操作，但都不起作用，并给出了错误：

    - name: Print found entries
      debug:
        msg: "{{ item.split()[1] }}"
      with_items: "{{ target.split(',') }}"

也尝试以下操作：

 with_items: "{{ target.results }}"
 with_items: "{{ target.stdout_lines }}"
 with_items: "{{ target.stdout }}"

收到错误：

TASK [debug] *******************************************************************
task path: /app/test.yml:917
fatal: [10.9.9.11]: FAILED! => {
    "msg": "The task includes an option with an undefined variable. The error was: 'list object' has no attribute 'split'\n\nThe error appears to be in '/app/test.yml': line 917, column 7, but may\nbe elsewhere in the file depending on the exact syntax problem.\n\nThe offending line appears to be:\n\n\n    - debug:\n      ^ here\n"
}

Answer 1

变量target是一个列表。可以迭代它。任务

    - debug:
        msg: "{{ item.split()[1] }}"
      loop: "{{ target }}"

给予
    "msg": "/web/test1.crt"
    "msg": "/web/SSL.crt"
    "msg": "/web/test.crt"