您如何比较两个可取列表中的值？

Question

我在Ansible中有两个列表，可以使用查找模块

- name: "Find all files in {{ backup_path }}"
  ansible.builtin.find:
    paths:
      - "{{ backup_path }}"
    get_checksum: true
    recurse: yes
    depth: 2
    patterns: "{{ file_patterns }}"
  register: backup_files

- name: "Find all files in {{ new_path }}"
  ansible.builtin.find:
    paths:
      - "{{ new_path }}"
    get_checksum: true
    recurse: yes
    depth: 2
    patterns: "{{ file_patterns }}"
  register: new_files

，一旦我注册这些变量，我想在old_files和new_files 列表。

- name: diff files between new and old
  ansible.builtin.shell: 
    cmd: "diff {{ item.0 }} {{ item.1 }}"
  register: diff_files
  loop: 
    - "{{ backup_files }}"
    - "{{ new_files }}"

我知道上面的片段是错误的，但是我想将0个项目从new_files与old_files n次数的0次数进行比较。

本剧本的目的是比较两个目录中的所有文件，并替换那些不同的文件。

Answer 1

事实证明，我使这比需要的要复杂得多，这是更简单的工作。基本上，使用diff -rq \ path1 \ path2。

- name: diff conf files between new and old
  ansible.builtin.shell: 
    cmd: "diff -rq {{ new_path }}conf {{ backup_path }}conf"
  register: diff_conf_files
  failed_when: diff_conf_files.rc not in [0,1]
  ignore_errors: yes

- name: diff bin files between new and old
  ansible.builtin.shell: 
    cmd: "diff -rq {{ new_path }}bin {{ backup_path }}bin"
  register: diff_bin_files
  failed_when: diff_bin_files.rc not in [0,1]
  ignore_errors: yes