如何在 Ansible 中对版本号进行排序 [英] How to sort version numbers in Ansible

问题描述

我正在构建一个 Ansible playbook，我想在其中备份数据库，以防我需要升级软件.为此，我想将可用的最高版本号与已安装的版本号进行比较.如果最新版本比安装版本高，我会备份数据库.

I'm building an Ansible playbook in which I want to make a backup of a database in case I need to upgrade the software. For this I want to compare the highest version number that is available to the version number that is installed. In case the latest version is hight than the installed version I'll back up the database.

然而，问题是我找不到在 Ansible 中对版本号进行排序的好方法.标准排序过滤器对字符串而不是数字/版本进行排序.

The problem however is that I cannot find a good way to sort version numbers in Ansible. The standard sort filter sorts on strings instead of numbers/versions.

这就是我现在正在做的:

This is what I'm doing right now:

- name: Get package version
  yum:
    list: package
  register: software_version

- name: Read which version is installed and available
  set_fact:
    software_version_installed: "{{ software_version | json_query(\"results[?yumstate=='installed'].version\") | sort | last }}"
    software_version_available: "{{ software_version | json_query(\"results[?yumstate=='available'].version\") | sort | last }}"

- name: Backup old database file on remote host
  copy:
    src: "{{ software.database_path }}"
    dest: "{{ software.database_path }}_{{ ansible_date_time.date }}_v{{ software_version_installed }}"
    remote_src: yes
  when: software_version_installed is version(software_version_available, "<")

只要版本号低于数字 10(例如 1.2.3，但不是 1.10.1)，上面的剧本就可以工作，因为排序就像字符串一样执行.当版本号必须排序时，例如1.2.3 和 1.10.1，以 1.2.3 为最新版本.

The playbook above works, as long as version numbers stay underneath the number 10 (e.g. 1.2.3, but not 1.10.1) since sorting is performed like a string. When the version number has to sort e.g. 1.2.3 and 1.10.1, it will take 1.2.3 as latest version.

显示问题:

- name: Read which version is installed and available
  set_fact:
    software_versions: [ "2.5.0", "2.9.0", "2.10.0", "2.11.0" ]

- name: Debug
  debug:
    var: software_versions | sort

TASK [grafana : Debug]**********************************
ok: [localhost] => {
    "software_versions | sort": [
        "2.10.0",
        "2.11.0",
        "2.5.0",
        "2.9.0"
    ]
}

有谁知道在 Ansible 中对版本号进行排序的好方法吗?

Does anyone know a good way to sort version numbers in Ansible?

推荐答案

问:有谁知道在 Ansible 中对版本号进行排序的好方法吗?

A:使用 filter_plugin.例如过滤器

A: Use filter_plugin. For example the filter

shell> cat filter_plugins/version_sort.py
from distutils.version import LooseVersion

def version_sort(l):
    return sorted(l, key=LooseVersion)

class FilterModule(object):

    def filters(self):
        return {
            'version_sort' : version_sort
            }

与剧本

shell> cat test-versions.yml
- name: Sort versions
  hosts: localhost

  vars:
    versions:
      - "0.1.0"
      - "0.1.5"
      - "0.11.11"
      - "0.9.11"
      - "0.9.3"
      - "0.10.2"
      - "0.6.1"
      - "0.6.0"
      - "0.11.0"
      - "0.6.5"
    
  tasks:
    - debug:
        msg: "{{ versions|version_sort }}"

给予

    "msg": [
        "0.1.0", 
        "0.1.5", 
        "0.6.0", 
        "0.6.1", 
        "0.6.5", 
        "0.9.3", 
        "0.9.11", 
        "0.10.2", 
        "0.11.0", 
        "0.11.11"
    ]

为方便起见，该过滤器可在 Github ansible-plugins 上找到.

For your convenience, the filter is available at Github ansible-plugins.

版本比较 完成迭代工作列表和比较项目.看下面的例子

Version comparison does the job to iterate the list and compare items. See the example below

shell> cat test-versions.yml
- hosts: localhost
    vars:
      version_installed: "1.10.1"
      versions:
        - "1.1.3"
        - "1.2.3"
        - "1.7.5"
        - "1.10.7"
    tasks:
      - debug: msg="{{ item }} is newer than {{ version_installed }}"
        loop: "{{ versions }}"
        when: item is version(version_installed, '>')

shell> ansible-playbook test-versions.yml | grep msg
"msg": "1.10.7 is newer than 1.10.1"

这篇关于如何在 Ansible 中对版本号进行排序的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！