Question

我需要找到未知目录位置的文件并将其删除。试图使用＆＃34;发现＆＃34;模块，注册其输出，并将其传递给＆＃34;文件＆＃34;。

即使我看到注册路径，我也不能在以后使用它：

< TASK [print find_result] >
ok: [1.2.3.4] => {
    "find_result": {
        "changed": false, 
        "examined": 3119, 
        "files": [
            {
                "atime": 1483973253.7295375, 
           ...
                "mode": "0600", 
                "mtime": 1483973253.7295375, 
                "nlink": 1, 
                "path": "/tmp/delme",

我的剧本是：

- hosts: "{{ target }}"
  become: no
  vars:
    find_what: "delme*"
    find_where: "/tmp"

  tasks:
  - name: finding files
    find:
      paths:            "{{ find_where }}"
      patterns:         "{{ find_what }}"
      recurse:          "yes"
      file_type:        "file"
    register: find_result

# \/ for debugging
  - name: print find_result
    debug: var=find_result

  - name: remove files
    file:
        path= "{{ item.path }}"
        state=absent
    with_items: "{{ find_result.files }}"

Answer 1

file之后的=任务空间存在语法缺陷。

尝试：

- name: remove files
  file:
    path: "{{ item.path }}"
    state: absent
  with_items: "{{ find_result.files }}"

Answer 2

删除文件任务时出现语法错误。

您可以使用：

- name: remove files
  file: >
    path={{ item.path }}
    state=absent
  with_items: "{{ find_result.files }}"

或者

- name: remove files
  file: path={{ item.path }} state=absent
  with_items: "{{ find_result.files }}"

Ansible - 如何注册来自＆＃34; FIND＆＃34;模块和其他用途

2 个答案: